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

Solving Trigonometric Equations Using Identities

B3.4 B3.3About 75 minutes
Learning goals
  • I can use a Pythagorean, double angle or compound angle identity to rewrite an equation in terms of one ratio of one angle.
  • I can choose the form of cos 2x that matches the rest of the equation.
  • I can solve the resulting linear or quadratic equation for 0 ≤ x ≤ 2π.
  • I can recognize when a step such as squaring may create extraneous solutions, and check my answers.

Words to know

extraneous solution
A value produced by the algebra (often after squaring both sides) that does not satisfy the original equation.
verify a solution
Substitute it back into the ORIGINAL equation, exactly or on a calculator, to check that both sides are equal.
single-ratio form
An equation written with only one trigonometric ratio of one angle, such as 2sin²x − 3 sin x + 1 = 0.

When the equation mixes things up

In the last lesson every equation had just one ratio, such as sin x. Real equations are messier: 2cos²x + 3 sin x − 3 = 0 has both cos x and sin x, and sin 2x = cos x has two different angles. The fix is to use an identity from lessons 3 to 9 to rewrite the equation in terms of one ratio of one angle. Then it is a linear or quadratic equation you already know how to solve.

This is where proving identities (B3.3) and solving equations (B3.4) come together.

Two different ratios or angles?Use an identity to get ONE ratio of ONE anglecos²x→1 − sin²xsin²x→1 − cos²xsin 2x→2 sin x cos xcos 2x→1 − 2sin²x or 2cos²x − 1sec²x→1 + tan²xThen move everything to one side, factor, and solve each factor.
The identities that most often turn a mixed equation into a single-ratio equation.
Worked example

Example 1: A Pythagorean substitution

Solve 2cos²x + 3 sin x − 3 = 0 for 0 ≤ x ≤ 2π.

  1. Two ratios. Replace cos²x with 1 − sin²x: 2(1 − sin²x) + 3 sin x − 3 = 0.
  2. Simplify: −2sin²x + 3 sin x − 1 = 0. Multiply by −1: 2sin²x − 3 sin x + 1 = 0.
  3. Factor: (2 sin x − 1)(sin x − 1) = 0.
  4. sin x = 1/2: x = π/6, 5π/6. sin x = 1: x = π/2.

x = π/6, π/2, 5π/6

Worked example

Example 2: A double angle with a common factor

Solve sin 2x = cos x for 0 ≤ x ≤ 2π.

  1. Two angles. Expand: 2 sin x cos x = cos x.
  2. Do not divide by cos x. Move everything to one side: 2 sin x cos x − cos x = 0.
  3. Factor: cos x(2 sin x − 1) = 0.
  4. cos x = 0: x = π/2, 3π/2. sin x = 1/2: x = π/6, 5π/6.

x = π/6, π/2, 5π/6, 3π/2

Worked example

Example 3: Choosing the form of cos 2x

Solve cos 2x = sin x for 0 ≤ x ≤ 2π.

  1. The right side has sin x, so use cos 2x = 1 − 2sin²x: 1 − 2sin²x = sin x.
  2. Rearrange: 2sin²x + sin x − 1 = 0, which factors as (2 sin x − 1)(sin x + 1) = 0.
  3. sin x = 1/2: x = π/6, 5π/6. sin x = −1: x = 3π/2.

x = π/6, 5π/6, 3π/2

xπ/2π3π/22π1−1π/65π/63π/2solid: y = cos 2x; dashed: y = sin x
The curves y = cos 2x and y = sin x meet at π/6, 5π/6 and 3π/2, the three solutions of Example 3. At 3π/2 they just touch, both equal to −1.
Worked example

Example 4: The cosine form

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

  1. The other term is cos x, so use cos 2x = 2cos²x − 1: 2cos²x − 1 + cos x = 0.
  2. 2cos²x + cos x − 1 = 0, which factors as (2 cos x − 1)(cos x + 1) = 0.
  3. cos x = 1/2: x = π/3, 5π/3. cos x = −1: x = π.

x = π/3, π, 5π/3

Worked example

Example 5: Secant and tangent

Solve sec²x − tan x − 1 = 0 for 0 ≤ x ≤ 2π.

  1. Replace sec²x with 1 + tan²x: 1 + tan²x − tan x − 1 = 0, so tan²x − tan x = 0.
  2. Factor: tan x(tan x − 1) = 0.
  3. tan x = 0: x = 0, π, 2π. tan x = 1: x = π/4, 5π/4.
  4. None of these makes cos x = 0, so sec x and tan x are defined at all of them ✓.

x = 0, π/4, π, 5π/4, 2π

Worked example

Example 6: A compound angle (and a warning about squaring)

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

  1. Method 1 (compound angle). √2 sin(x + π/4) = √2[sin x (√2/2) + cos x (√2/2)] = sin x + cos x. So the equation is √2 sin(x + π/4) = 1, or sin(x + π/4) = √2/2.
  2. Let θ = x + π/4, with π/4 ≤ θ ≤ 9π/4. sin θ = √2/2 at θ = π/4, 3π/4, 9π/4. So x = 0, π/2, 2π.
  3. Method 2 (squaring). (sin x + cos x)² = 1 gives 1 + sin 2x = 1, so sin 2x = 0 and x = 0, π/2, π, 3π/2, 2π.
  4. Check in the ORIGINAL equation: at π, sin π + cos π = −1 ≠ 1; at 3π/2, −1 + 0 = −1 ≠ 1. These are extraneous: squaring made −1 look like 1.

x = 0, π/2, 2π. If you square, always check every answer.

Always check

Substituting each answer back into the original equation takes a minute and catches lost signs, extraneous solutions from squaring, and values where tan x or sec x is undefined. A quick graph of both sides (radian mode) also shows how many solutions to expect.

Just for fun

Riddle: What did sin 2x say to cos x?

"Let's factor in what we have in common." (2 sin x cos x − cos x = cos x(2 sin x − 1).)

Hands-on activity

Match the strategy

You need: 12 index cards (or squares of paper) and a pencil.

Step 1. On six cards write these equations: (A) 2sin²x + 3 cos x − 3 = 0, (B) sin 2x = sin x, (C) cos 2x = cos x, (D) sec²x = 2 tan x, (E) cos 2x + 3 sin x − 2 = 0, (F) tan x = 2 sin x.

Step 2. On six more cards write these first moves: (1) sin²x → 1 − cos²x, (2) sin 2x → 2 sin x cos x, (3) cos 2x → 2cos²x − 1, (4) sec²x → 1 + tan²x, (5) cos 2x → 1 − 2sin²x, (6) tan x → sin x/cos x.

Step 3. Shuffle and match each equation with its first move. Then solve two of them fully and check them against the practice answers below.

What to look for: A–1, B–2, C–3, D–4, E–5, F–6. Equations A, B, C, E and F are solved in Your turn; D is in On your own.

Your turn

1

Solve 2sin²x + 3 cos x − 3 = 0 for 0 ≤ x ≤ 2π.

2

Solve sin 2x = sin x for 0 ≤ x ≤ 2π.

3

Solve cos 2x = cos x for 0 ≤ x ≤ 2π.

4

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

5

Solve tan x = 2 sin x for 0 ≤ x ≤ 2π.

6

Solve 2cos²x − sin²x = 2 for 0 ≤ x ≤ 2π.

More practice: warm-up

1

Write cos²x in terms of sin x.

2

Write sin 2x in terms of sin x and cos x.

3

An equation has cos 2x and sin x. Which form of cos 2x should you use?

4

Write sec²x in terms of tan x.

5

Write 1 − cos²x as a single term.

More practice: on your own

1

Solve 3 cos x + 3 = 2sin²x for 0 ≤ x ≤ 2π.

2

Solve sin 2x + cos x = 0 for 0 ≤ x ≤ 2π.

3

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

4

Solve sec²x = 2 tan x for 0 ≤ x ≤ 2π.

5

Solve sin²x − cos²x = 1/2 for 0 ≤ x ≤ 2π.

Challenge

1

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

2

Solve sin 4x = sin 2x for 0 ≤ x ≤ 2π.

3

Solve cos x − sin x = 1 for 0 ≤ x ≤ 2π by squaring, and identify any extraneous solutions.

Check your understanding

Quiz

1.

What is the best first step to solve 2cos²x + 3 sin x − 3 = 0?

2.

To solve sin 2x = cos x, what should you do?

3.

Which form of cos 2x makes cos 2x = sin x easiest to solve?

4.

What are the solutions of cos 2x + cos x = 0 for 0 ≤ x ≤ 2π?

5.

Why must you check your answers after squaring both sides of an equation?

6.

Using an identity, sec²x − tan x − 1 = 0 becomes which equation?

For parents and teachers

Big idea (B3.4 with B3.3): when an equation mixes ratios (sin and cos) or angles (x and 2x), use an identity to rewrite it with one ratio of one angle. Then factor and solve as before. This is where the identities from earlier lessons pay off.

How to help: Ask, "Which identity makes everything the same?" before your teen starts writing. After they solve, have them check each answer in the original equation on a calculator (radian mode).

Common difficulties:

  • Choosing the wrong form of cos 2x, which leaves two ratios.
  • Dividing by sin x or cos x and losing solutions.
  • Sign errors when replacing cos²x with 1 − sin²x (distribute the coefficient).
  • Not checking for extraneous solutions after squaring.