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Mom's Basement University

Basic Trigonometric Equations

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to solve basic trigonometric equations involving sine, cosine, and tangent, including finding multiple solutions within one period.

The Concept

Section titled “The Concept”

To solve a trigonometric equation:

  1. Isolate the trigonometric function (get sin θ, cos θ, or tan θ by itself)
  2. Use the unit circle or inverse functions to find the reference angle
  3. Find all solutions within one period (0 to 2π for sin/cos, 0 to π for tan)
  4. For general solutions, add the period multiplied by any integer k

General solution patterns:

  • sin θ = k has solutions at θ = arcsin(k) and θ = π - arcsin(k), plus 2πn
  • cos θ = k has solutions at θ = ±arccos(k), plus 2πn
  • tan θ = k has solutions at θ = arctan(k), plus πn

Always check solutions in the original equation, especially if you squared both sides or used identities that could introduce extraneous solutions.

Worked Example

Section titled “Worked Example”

1. Solve sin θ = 1/2 for θ in [0, 2π)

From the unit circle, sin θ = 1/2 at two angles in [0, 2π):

θ=π6orθ=π−π6=5π6\theta = \frac{\pi}{6} \quad \text{or} \quad \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

The first is in Q1 (where sin is positive), the second is in Q2 (where sin is also positive).

2. Solve 2 cos θ - 1 = 0 for θ in [0, 2π)

First isolate cos θ:

2cos⁡θ=1⇒cos⁡θ=122\cos\theta = 1 \quad \Rightarrow \quad \cos\theta = \frac{1}{2}

From the unit circle, cos θ = 1/2 at:

θ=π3orθ=2π−π3=5π3\theta = \frac{\pi}{3} \quad \text{or} \quad \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

Q1 and Q4, where cosine is positive.

3. Solve tan θ = 1 for θ in [0, 2π)

tan θ = 1 when sin θ and cos θ are equal. From the unit circle:

θ=π4orθ=π4+π=5π4\theta = \frac{\pi}{4} \quad \text{or} \quad \theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}

Tangent repeats every π, so we get one solution in Q1 and one in Q3.

4. Solve sin(2θ) = √3/2 for θ in [0, 2π)

Let u = 2θ. Then sin(u) = √3/2, which gives:

u=π3oru=2π3u = \frac{\pi}{3} \quad \text{or} \quad u = \frac{2\pi}{3}

Since u = 2θ and we need θ in [0, 2π), u ranges from 0 to 4π. Adding 2π to each:

u=π3,  2π3,  7π3,  8π3u = \frac{\pi}{3},\; \frac{2\pi}{3},\; \frac{7\pi}{3},\; \frac{8\pi}{3}

Divide by 2 to get θ:

θ=π6,  π3,  7π6,  4π3\theta = \frac{\pi}{6},\; \frac{\pi}{3},\; \frac{7\pi}{6},\; \frac{4\pi}{3}

Real-World Application

Section titled “Real-World Application”

Trigonometric equations are used to solve:

  • Finding angles in construction and engineering
  • Determining times of day for certain lighting or tidal conditions
  • Solving for angles in physics problems (projectile motion, waves)
  • Navigation and surveying problems

Example: finding the angle a ladder should be placed against a wall for safety involves solving a trigonometric equation.

Quiz

Section titled “Quiz”
Solve $\sin\theta = \frac{1}{2}$ for $\theta$ in $[0, 2\pi)$:
The general solution for $\tan\theta = k$ adds multiples of:
Solve $\cos\theta = 0$ for $\theta$ in $[0, 2\pi)$:
Why is it important to check solutions in trig equations?
Solve $2\sin\theta = \sqrt{2}$ for $\theta$ in $[0, 2\pi)$: