Grade by Grade
B. Trigonometric Functions · Trigonometric Identities and Equations

Solving Linear Trigonometric Equations in Radians

B3.4About 70 minutes
Learning goals
  • I can isolate the trigonometric ratio in a linear equation such as 3 tan x + 4 = 0.
  • I can find the related acute angle α exactly or with a calculator in radian mode.
  • I can use the CAST rule to find every solution in the interval 0 ≤ x ≤ 2π, including endpoints.
  • I can recognize equations with no solution, and check my solutions with a graph.

Words to know

linear trigonometric equation
An equation in which the trigonometric ratio appears only to the first power, such as 2 sin x − 1 = 0.
related acute angle (reference angle)
The acute angle α between the terminal arm and the x-axis, with 0 ≤ α ≤ π/2. It is found from the absolute value of the ratio.
CAST rule
A memory aid for which ratios are positive in each quadrant: All in I, Sine in II, Tangent in III, Cosine in IV.
solution set
All values of x in the given interval that make the equation true.
inverse trigonometric function
sin⁻¹, cos⁻¹ and tan⁻¹ on a calculator. Each returns only ONE angle (sin⁻¹ and tan⁻¹ between −π/2 and π/2, cos⁻¹ between 0 and π).

From proving to solving

Lessons 7 to 9 were about identities, which are true for every x. Now we turn to conditional equations, which are true only for certain x, and our job is to find them all. The curriculum asks for solutions "for the domain of real values from 0 to 2π", so in every equation in this unit, x is in 0 ≤ x ≤ 2π unless stated otherwise.

Remember from Grade 11: you solved equations like sin θ = 0.4 for 0° ≤ θ ≤ 360° and found two angles with the same ratio. Now we do the same thing in radians, and with the ratio buried inside an equation. Put your calculator in radian mode.

A five-step method

  1. Isolate the ratio: get sin x, cos x or tan x alone on one side.
  2. Check that the value is possible: sin x and cos x must be between −1 and 1. (Tangent can be any number.)
  3. Find α, the related acute angle, from the absolute value of the ratio: exactly (special angles) or with sin⁻¹, cos⁻¹ or tan⁻¹ in radian mode.
  4. Choose quadrants using the sign of the ratio and CAST.
  5. Write the solutions from α: quadrant I: α; quadrant II: π − α; quadrant III: π + α; quadrant IV: 2π − α. If the value is 0, 1 or −1, read the answer off the axes instead (and remember both endpoints 0 and 2π).
απ − απ + α2π − αA: all positiveS: sin positiveT: tan positiveC: cos positive0, 2πππ/23π/2
One related acute angle α gives four angles, one in each quadrant. CAST tells you which ones have the sign you need.
Worked example

Example 1: An exact solution

Solve 2 sin x − 1 = 0 for 0 ≤ x ≤ 2π.

  1. Isolate: 2 sin x = 1, so sin x = 1/2.
  2. α: sin α = 1/2 gives α = π/6.
  3. Sine is positive in quadrants I and II.
  4. x = π/6 or x = π − π/6 = 5π/6.

x = π/6 or 5π/6

xπ/2π3π/22π11/2−1y = 1/2π/65π/6y = sin x
The line y = 1/2 crosses one period of y = sin x exactly twice, at π/6 and 5π/6. A graph is a good check on the number of solutions.
Worked example

Example 2: A negative value

Solve √2 cos x + 1 = 0 for 0 ≤ x ≤ 2π.

  1. cos x = −1/√2 = −√2/2.
  2. α = π/4 (from cos α = √2/2; use the absolute value).
  3. Cosine is negative in quadrants II and III.
  4. x = π − π/4 = 3π/4 and x = π + π/4 = 5π/4.

x = 3π/4 or 5π/4

Worked example

Example 3: Using a calculator

Solve 3 tan x + 4 = 0 for 0 ≤ x ≤ 2π. Give answers to two decimal places.

  1. tan x = −4/3.
  2. α = tan⁻¹(4/3) ≈ 0.9273 (radian mode, positive value).
  3. Tangent is negative in quadrants II and IV.
  4. x = π − 0.9273 ≈ 2.2143 and x = 2π − 0.9273 ≈ 5.3559.
  5. Check: tan(2.21) ≈ −1.33 ✓.

x ≈ 2.21 or 5.36

Worked example

Example 4: The calculator gives only one answer

Solve 5 cos x = 2 for 0 ≤ x ≤ 2π, to two decimal places.

  1. cos x = 0.4.
  2. α = cos⁻¹(0.4) ≈ 1.1593.
  3. Cosine is positive in quadrants I and IV.
  4. x ≈ 1.1593 and x = 2π − 1.1593 ≈ 5.1239.

x ≈ 1.16 or 5.12

Worked example

Example 5: Reciprocal ratios and axis values

Solve (a) csc x = −2 and (b) 4 sin x + 4 = 0 for 0 ≤ x ≤ 2π.

  1. (a) csc x = −2 means sin x = −1/2. α = π/6, sine negative in quadrants III and IV: x = 7π/6 and 11π/6.
  2. (b) sin x = −1. This is an axis value: the point (0, −1) is at x = 3π/2 only.

(a) x = 7π/6, 11π/6; (b) x = 3π/2

Worked example

Example 6: No solution

Solve 2 sin x + 5 = 1.

  1. 2 sin x = −4, so sin x = −2.
  2. Sine is never less than −1.

No solution.

Endpoints count

The interval 0 ≤ x ≤ 2π includes both ends. So sin x = 0 has three solutions: 0, π and 2π. And cos x = 1 has two: 0 and 2π. Read the question carefully: if it says 0 ≤ x < 2π, then 2π is left out.

Just for fun

Your calculator only ever gives one answer to sin⁻¹. It is not being lazy: inverse sine is defined to return a value between −π/2 and π/2, so it can only report one angle. Finding the other answers is your job (and CAST is your assistant).

Hands-on activity

Find the second answer with a ruler

You need: a large circle (trace a dinner plate or a big lid) on paper, a ruler, a protractor, a pencil and a calculator.

Step 1. Draw axes through the centre. Measure the radius r.

Step 2. To solve sin x = 0.6, draw a horizontal line at height 0.6r above the x-axis. It crosses the circle at two points.

Step 3. Draw both terminal arms and measure both angles from the positive x-axis with the protractor. Convert to radians (× π/180).

Step 4. Repeat for cos x = −0.5 with a vertical line at 0.5r to the left of the y-axis.

What to look for: for sin x = 0.6 you should get about 37° and 143° (0.64 and 2.50 rad). For cos x = −0.5 you should get 120° and 240° (2π/3 ≈ 2.09 and 4π/3 ≈ 4.19 rad). The two answers in each case are mirror images, which is why one α gives both.

Your turn

1

Solve 2 cos x − √3 = 0 for 0 ≤ x ≤ 2π.

2

Solve √3 tan x + 1 = 0 for 0 ≤ x ≤ 2π.

3

Solve 4 sin x = −3 for 0 ≤ x ≤ 2π, to two decimal places.

4

Solve sec x = 2 for 0 ≤ x ≤ 2π.

5

Solve 3 cos x + 3 = 0 for 0 ≤ x ≤ 2π.

6

Solve 5 sin x − 2 = 1 for 0 ≤ x ≤ 2π, to two decimal places.

7

Solve sin x = 0 for 0 ≤ x ≤ 2π.

More practice: warm-up

1

Solve sin x = 1 for 0 ≤ x ≤ 2π.

2

Solve cos x = 0 for 0 ≤ x ≤ 2π.

3

Solve tan x = 1 for 0 ≤ x ≤ 2π.

4

Solve cos x = 1/2 for 0 ≤ x ≤ 2π.

5

Solve sin x = −√2/2 for 0 ≤ x ≤ 2π.

More practice: on your own

1

Solve 2 sin x + √3 = 0 for 0 ≤ x ≤ 2π.

2

Solve 7 tan x − 2 = 3 tan x + 6 for 0 ≤ x ≤ 2π, to two decimal places.

3

Solve csc x = −3 for 0 ≤ x ≤ 2π, to two decimal places.

4

Solve cot x + 1 = 0 for 0 ≤ x ≤ 2π.

5

Solve 2 cos x − 3 = 0. Explain your answer.

Challenge

1

Solve sin x = cos x for 0 ≤ x ≤ 2π. Explain why it is safe to divide by cos x.

2

Solve |2 sin x| = 1 for 0 ≤ x ≤ 2π.

3

For which values of k does 3 cos x = k have exactly one solution with 0 ≤ x ≤ 2π? Explain.

Check your understanding

Quiz

1.

What are the solutions of 2 sin x = 1 for 0 ≤ x ≤ 2π?

2.

In which quadrants are the solutions of tan x = −2?

3.

How many solutions does sin x = 0 have for 0 ≤ x ≤ 2π?

4.

If α is the related acute angle, which expression gives the solution in quadrant IV?

5.

How many solutions does 4 cos x = 6 have?

6.

To solve 5 cos x = 2 you press cos⁻¹(0.4) and get 1.1593. What is the other solution in 0 ≤ x ≤ 2π?

For parents and teachers

Big idea (B3.4): to solve a linear trigonometric equation, isolate the ratio, find the related acute angle α, then use CAST to find every solution between 0 and 2π. A calculator gives only one angle; the others come from α.

How to help: Ask your teen to sketch the four-quadrant diagram (α, π − α, π + α, 2π − α) before each question. Check that the calculator is in radian mode: a classic mistake gives 30 instead of 0.52.

Common difficulties:

  • Using the negative value in the inverse key, which gives an angle outside the interval; use the absolute value to find α.
  • Finding only one solution.
  • Forgetting the endpoint 2π (sin x = 0 has three solutions).
  • Not noticing an impossible value like sin x = −2.