Basic Trigonometric Equations

A basic trigonometric equation involves a single trigonometric function equal to a constant, such as $\sin x = a$, $\cos x = a$, or $\tan x = a$. To solve these equations, we use the unit circle and our knowledge of reference angles.

The Unit Circle and Principal Solutions

When you solve an equation like $\cos x = \frac{\sqrt{2}}{2}$, you are looking for the angles $x$ whose cosine (the x-coordinate on the unit circle) is equal to $\frac{\sqrt{2}}{2}$.

Because the unit circle is symmetric, there are usually two angles within a single full rotation, $[0, 2\pi)$, that satisfy the equation (unless the constant is $1$, $-1$, or $0$).

To find them:

1. Find the reference angle in the first quadrant.
2. Use the sign of the constant (positive or negative) to determine which quadrants the solutions lie in.

Finding General Solutions

Trigonometric functions are periodic, meaning their values repeat infinitely. If a problem asks for all solutions (not just those in a specific interval), you must write a general solution by adding multiples of the function's period.

Sine and Cosine

The period of both sine and cosine is $2\pi$. If $x_1$ and $x_2$ are your principal solutions in $[0, 2\pi)$, the general solutions are: $x = x_1 + 2k\pi \quad \text{and} \quad x = x_2 + 2k\pi$ where $k$ is any integer ($k \in \mathbb{Z}$).

Tangent

The period of the tangent function is $\pi$. Because tangent values repeat every half-circle, you only need to find the principal solution $x_1$ in the interval $[0, \pi)$ or $(-\frac{\pi}{2}, \frac{\pi}{2})$ and add multiples of $\pi$: $x = x_1 + k\pi$

Example 1: Solving in a Specific Interval

Problem: Find all solutions of $\sin x = \frac{1}{2}$ in $[0, 2\pi)$.

Step 1: Identify the reference angle. We know from basic trig values that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Step 2: Determine the quadrants. The value $\frac{1}{2}$ is positive, and sine is positive in Quadrants I and II.

Step 3: Find the angles in those quadrants.

• Quadrant I: $x = \frac{\pi}{6}$
• Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}$

Example 2: Finding All Solutions (General Solution)

Problem: Solve $\tan x = -\sqrt{3}$ for all $x$.

Step 1: Identify the reference angle. We know $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

Step 2: Determine the quadrant for the principal solution. Tangent is negative in Quadrants II and IV. Let's find the angle in Quadrant II: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

Step 3: Write the general solution. Since the period of tangent is $\pi$, we just add $k\pi$ to our principal solution.

Answer: $x = \frac{2\pi}{3} + k\pi$, where $k$ is any integer.