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Lesson 2: Solving Equations Using Identities

Estimated time: 35-40 minutes

Learning Objectives

Using Pythagorean Identities

When an equation contains two different trig functions, use an identity to rewrite in terms of one function, then solve the resulting equation.

Example 1: Using sin² + cos² = 1

Solve 2sin²θ + cosθ − 1 = 0 on [0, 2π).

Step 1: Replace sin²θ = 1 − cos²θ: 2(1−cos²θ) + cosθ − 1 = 0

Step 2: Simplify: −2cos²θ + cosθ + 1 = 0, or 2cos²θ − cosθ − 1 = 0

Step 3: Factor: (2cosθ + 1)(cosθ − 1) = 0

Step 4: cosθ = −1/2 ⇒ θ = 2π/3, 4π/3. cosθ = 1 ⇒ θ = 0.

Solutions: {0, 2π/3, 4π/3}

Using Double-Angle Identities

Example 2: Double Angle

Solve cos2θ + cosθ = 0 on [0, 2π).

Replace cos2θ = 2cos²θ − 1: (2cos²θ−1) + cosθ = 0

2cos²θ + cosθ − 1 = 0 ⇒ (2cosθ−1)(cosθ+1) = 0

cosθ = 1/2 ⇒ θ = π/3, 5π/3. cosθ = −1 ⇒ θ = π.

Solutions: {π/3, π, 5π/3}

Example 3: sin2θ = cosθ

2sinθcosθ = cosθ ⇒ cosθ(2sinθ−1) = 0

cosθ = 0 ⇒ θ = π/2, 3π/2. sinθ = 1/2 ⇒ θ = π/6, 5π/6.

Solutions: {π/6, π/2, 5π/6, 3π/2}

Common Mistakes

Check Your Understanding

1. Solve 2cos²θ + sinθ = 1 on [0, 2π).

Replace cos²=1−sin²: 2(1−sin²)+sin−1=0 ⇒ −2sin²+sin+1=0 ⇒ 2sin²−sin−1=0 ⇒ (2sin+1)(sin−1)=0. sin=−1/2: θ=7π/6, 11π/6. sin=1: θ=π/2. {π/2, 7π/6, 11π/6}.

2. Solve sin2θ = sinθ on [0, 2π).

2sinθcosθ−sinθ=0 ⇒ sinθ(2cosθ−1)=0. sin=0: θ=0,π. cos=1/2: θ=π/3,5π/3. {0, π/3, π, 5π/3}.

3. Solve tan²θ − 1 = 0 on [0, 2π).

tanθ=±1: {π/4, 3π/4, 5π/4, 7π/4}.

Key Takeaways

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