Chapter 4 of 8

All 8 chapters
  1. 1Radians, readunderstood
  2. 2The unit circle and any angle, readunderstood
  3. 3Trig graphs and their changes, readunderstood
  4. 4Solving trig equations, readunderstood
  5. 5Identities, readunderstood
  6. 6Reciprocal and inverse trig functions, readunderstood
  7. 7Compound and double angles, readunderstood
  8. 8Combining sin and cos, readunderstood

M5.3.4 Chapter 4

Solving trig equations

If sin x = 0.4, what is x, and why are there so many answers?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve sin x = 0.4 for 0 ≤ x ≤ 2π
  • Solve 2cos²x − cos x − 1 = 0

An equation such as sin x = 0.4 asks: which angles have a sine of 0.4? On the unit circle, sin x is the height of the point. Draw the line at height 0.4, and it cuts the circle in two places. Turn full circles and you find more. So the equation has many solutions, and a question usually gives an interval to keep the list short.

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. A wave named sin x crosses a dashed line at height 0.4 twice. Dots with dashed guides to the x axis mark the two crossing points, labelled 0.412 and 2.730.0π/2π3π/22π−101x (radians)y0.4122.730sin x
The line y = 0.4 cuts y = sin x at two points between 0 and 2π. They are mirror images in the line x = π/2.

Draw it first

Draw the graph and the line

  1. A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. In step 1 the wave y = sin x appears. In step 2 a dashed line at height 0.4 appears. In step 3 two dots with dashed guide lines to the x axis mark where the line cuts the wave.0π/2π3π/22π−101x (radians)y

    1 Draw the wave y = sin x from 0 to 2π.

  2. A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. In step 1 the wave y = sin x appears. In step 2 a dashed line at height 0.4 appears. In step 3 two dots with dashed guide lines to the x axis mark where the line cuts the wave.0π/2π3π/22π−101x (radians)y

    2 Draw the horizontal line y = 0.4 across it.

  3. A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. In step 1 the wave y = sin x appears. In step 2 a dashed line at height 0.4 appears. In step 3 two dots with dashed guide lines to the x axis mark where the line cuts the wave.0π/2π3π/22π−101x (radians)y

    3 Mark the two places where the line cuts the wave. Read down to the x axis to see the two solutions.

Here is the method. Step 1: get one trig function on its own, equal to a number. Step 2: use the calculator for the acute reference angle α, from the size of the number, ignoring any minus sign. Step 3: use the quadrants where the function has the right sign. Step 4: add or take away full turns until every solution in the interval is found.

Set the calculator to the right mode first: radians if the interval uses π, degrees if it uses degree signs. The key marked sin⁻¹ or arcsin gives the angle for a given sine. It gives the one in the first quadrant, or a negative angle when the number is negative. We use it only for α.

A table with the heads equation and solutions. The rows read sin x = k with α, π − α. sin x = −k with π + α, 2π − α. cos x = k with α, 2π − α. cos x = −k with π − α, π + α. tan x = k with α, π + α. tan x = −k with π − α, 2π − α.equationsolutionssin x = kα, π − αsin x = −kπ + α, 2π − αcos x = kα, 2π − αcos x = −kπ − α, π + αtan x = kα, π + αtan x = −kπ − α, 2π − α
Solutions for 0 ≤ x ≤ 2π, with k above 0 (and at most 1 for sin and cos) and α the acute reference angle. The quadrants are those where the function has that sign. Use the same rows in degrees, with 180° for π and 360° for 2π.

Worked example

Solve sin x = 0.4 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

  1. The calculator is in radian mode. sin⁻¹ 0.4 = 0.41152. This is the reference angle α.
  2. sin is positive in the first and second quadrants. In the first, x = α = 0.41152.
  3. In the second, x = π − α = 3.14159 − 0.41152 = 2.73007.
  4. The next solutions would be 0.4115 + 2π and 2.7301 + 2π, which are more than 2π. So there are only two.
  5. To 3 decimal places, x = 0.412 or x = 2.730.
  6. Check: sin 0.412 = 0.40044 and sin 2.730 = 0.40007. Both are close to 0.4, and the small difference is from rounding.

Answer: x = 0.412 or x = 2.730

The same reference angle gives the answer in the unit circle picture. The two points at height 0.4 are mirror images in the y axis, so their angles are α and π − α.

A circle with centre O and axes named x and y. Two points named P and Q stand at the same height above the x axis, one in the first quadrant and one in the second. Dashed lines from each point run to both axes.xyPQO
P has angle α and Q has angle π − α. Both have height sin x = 0.4.

Worked example

Solve tan x = −2 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

  1. The calculator gives tan⁻¹(−2) = −1.10715. This is a negative angle, so it is outside the interval.
  2. The reference angle is α = 1.10715, from the size of the number 2.
  3. tan is negative in the second and fourth quadrants. In the second, x = π − α = 3.14159 − 1.10715 = 2.03444.
  4. In the fourth, x = 2π − α = 6.28319 − 1.10715 = 5.17604.
  5. To 3 decimal places, x = 2.034 or x = 5.176.

Answer: x = 2.034 or x = 5.176

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −4 to 4 named y. Dashed vertical lines stand at π/2 and 3π/2. A graph named tan x has three branches. A dashed line at height −2 meets the graph at two points marked with dots, at x = 2.034 and x = 5.176.0π/2π3π/22π−4−2024x (radians)ytan x
tan x = −2 has two solutions between 0 and 2π, one in each of the second and fourth quadrants, at x = 2.034 and x = 5.176.

When the angle is 2x, 3x or x − 20°, first change the interval. If 0 ≤ x ≤ 2π and the equation has 2x, then 2x runs from 0 to 4π. Solve for 2x in that whole longer interval, so that you find every solution, and only then divide by 2.

Worked example

Solve sin 2x = 0.5 for 0 ≤ x ≤ 2π. Give exact answers.

  1. Let θ = 2x. Then 0 ≤ x ≤ 2π gives 0 ≤ θ ≤ 4π. We solve sin θ = 0.5 in that longer interval.
  2. sin π/6 = 0.5, so α = π/6. In the first two quadrants, θ = π/6 and θ = π − π/6 = 5π/6.
  3. Add one full turn, 2π, to each: θ = π/6 + 2π = 13π/6 and θ = 5π/6 + 2π = 17π/6.
  4. The next ones, 25π/6 and 29π/6, are bigger than 4π, so we stop.
  5. Divide each by 2: x = π/12, 5π/12, 13π/12 and 17π/12.
  6. Check x = 5π/12: 2x = 5π/6 and sin 5π/6 = 0.5. This agrees.

Answer: x = π/12, 5π/12, 13π/12 and 17π/12

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. A fast wave named sin 2x has two cycles. A dashed line at height 0.5 cuts it at four points, marked with dots.0π/2π3π/22π−101x (radians)ysin 2x
sin 2x = 0.5 has four solutions between 0 and 2π, because the graph has two cycles.

Worked example

Solve sin(x − 20°) = 0.6 for 0° ≤ x ≤ 360°. Give the answers to 1 decimal place.

  1. Let θ = x − 20°. Then 0° ≤ x ≤ 360° gives −20° ≤ θ ≤ 340°. We solve sin θ = 0.6 in that interval.
  2. The calculator is in degree mode. sin⁻¹ 0.6 = 36.87°. So in the first quadrant θ = 36.87°.
  3. In the second quadrant θ = 180° − 36.87° = 143.13°.
  4. Both lie between −20° and 340°. The next ones, 396.87° and 503.13°, are too big. The earlier ones, −323.13° and −216.87°, are too small.
  5. Add 20° to each: x = 36.87° + 20° = 56.87° and x = 143.13° + 20° = 163.13°.
  6. To 1 decimal place, x = 56.9° or x = 163.1°.
  7. Check x = 163.1°: x − 20° = 143.1°, and sin 143.1° = 0.6004. This agrees.

Answer: x = 56.9° or x = 163.1°

A graph with the x axis named x (degrees) from 0 to 360 and a y axis from −1.5 to 1.5 named y. A wave named sin(x − 20°) is the sine wave moved 20 degrees to the right. A dashed line at height 0.6 cuts it at two points, labelled 56.9 and 163.1.060120180240300360−101x (degrees)y56.9163.1sin(x − 20°)
The wave is y = sin x moved 20° to the right. The line y = 0.6 cuts it at x = 56.9° and x = 163.1°.

Many equations are quadratics in sin x or cos x. Treat sin x or cos x as one unknown. Move everything to one side, factorise, and solve each factor separately. Each factor gives an easy equation of the kind you already know how to solve. Always factorise: never divide away a trig function, because that can lose solutions.

Worked example

Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π. Give exact answers.

  1. Let c = cos x. Then the equation is 2c² − c − 1 = 0.
  2. Factorise: (2c + 1)(c − 1) = 0. Check by expanding: 2c² − 2c + c − 1 = 2c² − c − 1.
  3. So 2c + 1 = 0 or c − 1 = 0. That gives cos x = −1/2 or cos x = 1.
  4. For cos x = −1/2: the reference angle is π/3. cos is negative in the second and third quadrants, so x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.
  5. For cos x = 1: the graph touches the top at x = 0 and x = 2π. Both ends are inside the interval.
  6. So x = 0, 2π/3, 4π/3 or 2π.

Answer: x = 0, 2π/3, 4π/3 or 2π

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −2 to 4 named y. A curve named f(x), which is 2cos²x − cos x − 1, starts at height 0 at x = 0, dips below the x axis between 0 and 2π/3, crosses at 2π/3, rises above the axis, falls to cross at 4π/3 and rises to 0 at 2π. Dots mark the four places where it meets the x axis.02π/34π/32π−2024x (radians)yf(x)
The curve y = 2cos²x − cos x − 1 (called f(x) here) meets the x axis at the four solutions: 0, 2π/3, 4π/3 and 2π.

General solutions give every solution at once, without an interval. Let n stand for any integer: positive, negative or 0. If sin x = sin α, then x = α + 2nπ or x = π − α + 2nπ. If cos x = cos α, then x = ±α + 2nπ. If tan x = tan α, then x = α + nπ. They are the quadrant rules with every full turn added: a full turn is 2π for sin and cos, but only π for tan.

Worked example

Find the general solution of 2cos²x + 3 sin x = 0, where n is any integer.

  1. Replace cos²x by 1 − sin²x. The equation becomes 2(1 − sin²x) + 3 sin x = 0.
  2. Expand: 2 − 2 sin²x + 3 sin x = 0. Move everything to one side: 2 sin²x − 3 sin x − 2 = 0.
  3. Factorise: (2 sin x + 1)(sin x − 2) = 0. Check: 2 sin²x − 4 sin x + sin x − 2 = 2 sin²x − 3 sin x − 2.
  4. So sin x = −1/2 or sin x = 2. The second has no solution, because sin x is never more than 1.
  5. sin x = −1/2: the reference angle is π/6. sin is negative in the third and fourth quadrants.
  6. So x = π + π/6 = 7π/6, or x = 2π − π/6, which is −π/6 + 2π.
  7. Add every full turn: x = −π/6 + 2nπ or x = 7π/6 + 2nπ.

Answer: x = −π/6 + 2nπ or x = 7π/6 + 2nπ

See it three ways

With things

A swing passes the same height twice on each swing, once going up and once coming down. If you are asked at which times the swing is 40 cm high, there are two answers for each full swing.

As a picture

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1.5 to 1.5 named y. A wave named sin x crosses a dashed line at height 0.5 twice. Dots with dashed guides mark the two crossing points, labelled π/6 and 5π/6.0π/2π3π/22π−101x (radians)yπ/65π/6sin x

In words

The first solution comes from the calculator. The second is its mirror image in the line x = π/2. Then every full turn gives more.

In symbols

sin x = 0.5: x = π/6 + 2nπ or x = 5π/6 + 2nπ, for n = 0, 1, 2, … and for negative n too.

Try it

Find the second solution on a graph

You need: squared paper, a ruler, a pencil.

  1. Draw y = sin x for x from 0° to 360°, plotting every 30°. Use 1 square for 0.1 up the side.
  2. Draw a horizontal line at height 0.5. Mark the places where the line cuts the wave.
  3. Read the two x values from the graph. They should be near 30° and 150°.
  4. Check each with a calculator: sin 30° and sin 150°.

What you will see: The two solutions are the same distance either side of 90°, the peak of the wave. They are 30° from 0° and 30° from 180°.

If you cannot try it, this is what happens.
A table with the heads x and sin x. The rows read 0°: 0. 30°: 0.5. 60°: 0.87. 90°: 1. 120°: 0.87. 150°: 0.5. 180°: 0.xsin x0°030°0.560°0.8790°1120°0.87150°0.5180°0

A wrong idea many people have

To solve 2 sin x cos x = cos x, divide both sides by cos x. Then 2 sin x = 1 gives x = π/6 and 5π/6.

Why it feels right: Dividing by the same thing on both sides is a normal step in algebra, and it removes a messy factor.

The test that breaks it: Try x = π/2. Then cos x = 0, so both sides are 0, and the equation holds. But the answers above do not include π/2. Dividing by cos x threw away every solution where cos x = 0.

What is true: Move everything to one side and factorise: cos x (2 sin x − 1) = 0. Then cos x = 0 gives x = π/2 and 3π/2, and sin x = 1/2 gives π/6 and 5π/6. There are four solutions.

A wrong idea many people have

sin x = 0.4 has the one solution x = 0.412 in 0 ≤ x ≤ 2π, because that is what the calculator says.

Why it feels right: The calculator gave an answer, and it looks complete.

The test that breaks it: Put x = 2.730 into the calculator: sin 2.730 = 0.4004. This is also a solution, and the calculator did not show it.

What is true: The key sin⁻¹ gives only an angle from −π/2 to π/2 (cos⁻¹ gives one from 0 to π). Use the graph or the unit circle to find the others. For sin and cos, expect two solutions in each full turn. For tan, expect one.

A wrong idea many people have

To solve sin 2x = 0.5 for 0 ≤ x ≤ 2π, find 2x = π/6 and then x = π/12 and 5π/12 only.

Why it feels right: The question is for x, so the interval seems to stay at 0 to 2π when we divide.

The test that breaks it: Check x = 13π/12. Then 2x = 13π/6, and sin 13π/6 = sin π/6 = 0.5. This x is in the interval, but the short list does not include it.

What is true: Change the interval for 2x first: 0 ≤ 2x ≤ 4π. Then find every solution for 2x in that interval and divide each by 2. This gives four solutions.

Key words

general solution
a formula, using an integer n, that gives every solution of a trig equation.
reference angle
the acute angle found from the size of the number, from which the solutions in every quadrant are built.

Back to your guess

The answer is C: There are two values of x between 0 and 2π, and the calculator gives only one.

Between 0 and 2π, the height 0.4 is reached twice, once going up and once coming down. The calculator gives only the first. We find the other from the symmetry of the graph, and the wave then repeats for ever.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. How many solutions does sin x = 0.7 have for 0 ≤ x ≤ 4π?

    Tap your answer.

    1. Not this one. That is the number between 0 and 2π. The interval here is twice as long, and the wave repeats. Go to M5.3.4. Try another.

    2. Yes, that is right. The line y = 0.7 cuts y = sin x twice in each full turn. The interval from 0 to 4π holds two full turns, so there are 2 × 2 = 4 solutions.

    3. Not this one. There are two solutions in each full turn of 2π, not four. The interval 0 to 4π holds two full turns. Go to M5.3.4. Try another.

    4. Not this one. The calculator gives one value, but the line at height 0.7 cuts the wave twice in every full turn. Go to M5.3.4. Try another.

    See the answer

    4

    The line y = 0.7 cuts y = sin x twice in each full turn. The interval from 0 to 4π holds two full turns, so there are 2 × 2 = 4 solutions.

    • If you chose A: That is the number between 0 and 2π. The interval here is twice as long, and the wave repeats. Go to M5.3.4.
    • If you chose C: There are two solutions in each full turn of 2π, not four. The interval 0 to 4π holds two full turns. Go to M5.3.4.
    • If you chose D: The calculator gives one value, but the line at height 0.7 cuts the wave twice in every full turn. Go to M5.3.4.

  2. The calculator gives cos⁻¹ 0.3 = 1.266 radians. What is the other solution of cos x = 0.3 for 0 ≤ x ≤ 2π?

    Tap your answer.

    1. Not this one. That is a correct solution of cos x = 0.3 but it is outside the interval 0 to 2π. Add 2π to bring it inside. Go to M5.3.4. Try another.

    2. Not this one. That is π − 1.266, which is the rule for sin x = 0.3, not for cos x = 0.3. cos is positive in the first and fourth quadrants. Go to M5.3.4. Try another.

    3. Not this one. That is π + 1.266, in the third quadrant. There cos is negative, so it is not a solution of cos x = 0.3. Go to M5.3.4. Try another.

    4. Yes, that is right. cos is positive in the first and fourth quadrants. In the fourth, x = 2π − α = 6.283 − 1.266 = 5.017. This is the same as −1.266 + 2π.

    See the answer

    5.017

    cos is positive in the first and fourth quadrants. In the fourth, x = 2π − α = 6.283 − 1.266 = 5.017. This is the same as −1.266 + 2π.

    • If you chose A: That is a correct solution of cos x = 0.3 but it is outside the interval 0 to 2π. Add 2π to bring it inside. Go to M5.3.4.
    • If you chose B: That is π − 1.266, which is the rule for sin x = 0.3, not for cos x = 0.3. cos is positive in the first and fourth quadrants. Go to M5.3.4.
    • If you chose C: That is π + 1.266, in the third quadrant. There cos is negative, so it is not a solution of cos x = 0.3. Go to M5.3.4.

  3. A learner solves sin x cos x = ½ sin x by dividing both sides by sin x. Which solutions in 0 ≤ x ≤ 2π are lost?

    Tap your answer.

    1. Yes, that is right. sin x (cos x − 1/2) = 0 gives sin x = 0 or cos x = 1/2. The first gives x = 0, π and 2π, and dividing by sin x removed them. The second gives π/3 and 5π/3.

    2. Not this one. At these values cos x = 0 and sin x = ±1, so the two sides are 0 and ±1/2. They are not solutions. Go to M5.3.4. Try another.

    3. Not this one. It is allowed only when the thing is not 0. sin x = 0 at x = 0, π and 2π, and those solutions are lost. Go to M5.3.4. Try another.

    4. Not this one. These come from cos x = 1/2, which the learner found. They are not lost. Go to M5.3.4. Try another.

    See the answer

    0, π and 2π, which are 0, 3.142 and 6.283

    sin x (cos x − 1/2) = 0 gives sin x = 0 or cos x = 1/2. The first gives x = 0, π and 2π, and dividing by sin x removed them. The second gives π/3 and 5π/3.

    • If you chose B: At these values cos x = 0 and sin x = ±1, so the two sides are 0 and ±1/2. They are not solutions. Go to M5.3.4.
    • If you chose C: It is allowed only when the thing is not 0. sin x = 0 at x = 0, π and 2π, and those solutions are lost. Go to M5.3.4.
    • If you chose D: These come from cos x = 1/2, which the learner found. They are not lost. Go to M5.3.4.

  4. Amina solves sin x = 0.5 for 0° ≤ x ≤ 360° and her calculator shows 0.5236. What should she do?

    Tap your answer.

    1. Not this one. The interval is in degrees, so she needs degree mode. Radian mode would give 0.5236 radians, which is 30°. Go to M5.3.1. Try another.

    2. Yes, that is right. The calculator was in radian mode: 0.5236 radians is 30°. In degree mode sin⁻¹ 0.5 = 30°. The second solution is 180° − 30° = 150°.

    3. Not this one. 0.5236 is the answer in radians. In the interval 0° to 360° it is only 0.5236°, which is too small to be a solution. Go to M5.3.1. Try another.

    See the answer

    Change to degree mode. Then the answers are 30° and 150°.

    The calculator was in radian mode: 0.5236 radians is 30°. In degree mode sin⁻¹ 0.5 = 30°. The second solution is 180° − 30° = 150°.

    • If you chose A: The interval is in degrees, so she needs degree mode. Radian mode would give 0.5236 radians, which is 30°. Go to M5.3.1.
    • If you chose C: 0.5236 is the answer in radians. In the interval 0° to 360° it is only 0.5236°, which is too small to be a solution. Go to M5.3.1.

  5. From an earlier chapter

    What is the period of y = sin 4x?

    Tap your answer.

    1. Not this one. You multiplied 2π by 4. The 4 makes the wave repeat four times as fast, so the period is divided by 4. Go to M5.3.3. Try another.

    2. Not this one. You divided π by 4. The period of sin x is 2π, not π. Divide 2π by 4. Go to M5.3.3. Try another.

    3. Yes, that is right. The period of sin bx is 2π ÷ b. With b = 4 the period is 2π/4 = π/2.

    4. Not this one. That is the period of sin 2x. The number next to x is 4, so divide 2π by 4. Go to M5.3.3. Try another.

    See the answer

    π/2

    The period of sin bx is 2π ÷ b. With b = 4 the period is 2π/4 = π/2.

    • If you chose A: You multiplied 2π by 4. The 4 makes the wave repeat four times as fast, so the period is divided by 4. Go to M5.3.3.
    • If you chose B: You divided π by 4. The period of sin x is 2π, not π. Divide 2π by 4. Go to M5.3.3.
    • If you chose D: That is the period of sin 2x. The number next to x is 4, so divide 2π by 4. Go to M5.3.3.

  6. Which pair of solutions does cos x = −1/2 have for 0 ≤ x ≤ 2π?

    Tap your answer.

    1. Not this one. These solve cos x = +1/2, with the first and fourth quadrants. Here cos is negative. Go to M5.3.4. Try another.

    2. Not this one. These are the angles with sin = 1/2. The reference angle for cos = 1/2 is π/3. Go to M5.3.4. Try another.

    3. Not this one. Both must be in the quadrants where cos is negative. 5π/3 is in the fourth quadrant, where cos is positive. Go to M5.3.4. Try another.

    4. Yes, that is right. The reference angle is π/3. cos is negative in the second and third quadrants, so x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.

    See the answer

    x = 2π/3 and 4π/3

    The reference angle is π/3. cos is negative in the second and third quadrants, so x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.

    • If you chose A: These solve cos x = +1/2, with the first and fourth quadrants. Here cos is negative. Go to M5.3.4.
    • If you chose B: These are the angles with sin = 1/2. The reference angle for cos = 1/2 is π/3. Go to M5.3.4.
    • If you chose C: Both must be in the quadrants where cos is negative. 5π/3 is in the fourth quadrant, where cos is positive. Go to M5.3.4.

  7. Which gives every solution of tan x = tan α, where n is any integer?

    Tap your answer.

    1. Yes, that is right. tan has period π, and it takes each value once in each period. So once we know one solution α, all the others are α + π, α + 2π, α − π, and so on: α + nπ.

    2. Not this one. That would include α + π/2, where tan is not defined or has a different value. tan repeats every π, not every π/2. Go to M5.3.4. Try another.

    3. Not this one. The solution is α itself, not twice α. Adding full periods of π to α gives all the others. Go to M5.3.4. Try another.

    4. Not this one. That is the pattern for sin or cos. tan repeats every π, so one value of α alone is repeated every π. Go to M5.3.4. Try another.

    See the answer

    x = α + nπ

    tan has period π, and it takes each value once in each period. So once we know one solution α, all the others are α + π, α + 2π, α − π, and so on: α + nπ.

    • If you chose B: That would include α + π/2, where tan is not defined or has a different value. tan repeats every π, not every π/2. Go to M5.3.4.
    • If you chose C: The solution is α itself, not twice α. Adding full periods of π to α gives all the others. Go to M5.3.4.
    • If you chose D: That is the pattern for sin or cos. tan repeats every π, so one value of α alone is repeated every π. Go to M5.3.4.

  8. Solve sin x = 0.3 for 0° ≤ x ≤ 360°. Give the answers to 1 decimal place.

    See the answer

    x = 17.5° or x = 162.5°

    In degree mode sin⁻¹ 0.3 = 17.46°, which is 17.5° to 1 decimal place. sin is also positive in the second quadrant. So x = 180° − 17.46° = 162.54°, which is 162.5°.

  9. Solve 3 sin 2x = 1 for 0 ≤ x ≤ π. Give the answers to 3 decimal places.

    See the answer

    x = 0.170 or x = 1.401

    Divide by 3: sin 2x = 1/3. The interval for 2x is 0 ≤ 2x ≤ 2π. In radian mode sin⁻¹(1/3) = 0.3398, so 2x = 0.3398 or 2x = π − 0.3398 = 2.8018. Halving gives x = 0.1699 and x = 1.4009, which are 0.170 and 1.401.

  10. Write the general solution of 2 cos x + 1 = 0, where n is any integer.

    See the answer

    x = 2π/3 + 2nπ or x = 4π/3 + 2nπ

    2 cos x + 1 = 0 gives cos x = −1/2. The reference angle is π/3, and cos is negative in the second and third quadrants. So x = π − π/3 = 2π/3 or x = π + π/3 = 4π/3. Add every full turn, 2nπ.

  11. Solve 4 cos x = 3 sin x for 0° ≤ x ≤ 360°. Give the answers to 1 decimal place.

    See the answer

    x = 53.1° or x = 233.1°

    If cos x = 0 then sin x = ±1, and 4 × 0 = ±3 is false. So cos x is not 0 and we may divide by cos x: 4 = 3 tan x, so tan x = 4/3. The calculator gives 53.13°. tan is also positive in the third quadrant, so x = 180° + 53.13° = 233.13°.

  12. Write the general solution of tan x = √3, where n is any integer.

    See the answer

    x = π/3 + nπ

    tan(π/3) = √3, so one solution is π/3. tan repeats every π, so every solution is π/3 plus a whole number of π: x = π/3 + nπ.

Keep it

Copy this into your exercise book.

Reference angle α from |k|. Choose quadrants by CAST. Add full turns. For sin 2x, double the interval first. Factorise, never divide by a trig function that can be 0.

Draw: A sine wave with a horizontal line cutting it twice, and the two solutions marked α and π − α.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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