9.5 Solving Trigonometric Equations

Solving Trigonometric Equations

Trigonometric equations ask you to find the angle(s) that make a trig statement true. Unlike regular algebra where you get a single answer, trig equations often produce infinitely many solutions because trig functions repeat. Knowing how to solve these equations is essential for applications in physics, engineering, and later math courses.

This section covers basic trig equations, quadratic forms, using identities to simplify, and multiple angle equations.

Solving Basic Trigonometric Equations

The core strategy is always the same: isolate the trig function, then figure out which angle(s) produce that value.

1. Use algebraic operations (adding, subtracting, multiplying, dividing) to get the trig function alone on one side.
2. Determine the reference angle using inverse trig functions or known values from the unit circle.
3. Identify all angles that work by checking which quadrants give the correct sign.
4. Write the general solution by adding the full period to account for repetition.

General solution formulas:

• For $\sin \theta = a$: solutions are $\theta = \arcsin(a) + 2\pi n$ or $\theta = \pi - \arcsin(a) + 2\pi n$, where $n$ is any integer
• For $\cos \theta = a$: solutions are $\theta = \arccos(a) + 2\pi n$ or $\theta = -\arccos(a) + 2\pi n$, where $n$ is any integer
• For $\tan \theta = a$: solutions are $\theta = \arctan(a) + \pi n$, where $n$ is any integer (tangent repeats every $\pi$, not $2\pi$)

If the problem specifies an interval like $[0, 2\pi)$, list only the solutions that fall within that interval instead of writing the general form.

Algebraic Techniques for Trigonometric Equations

Many trig equations require some algebra before you can read off the angle. Treat the trig function like a variable and solve for it first.

Example: Solve $2\sin \theta + 1 = 0$ on $[0, 2\pi)$.

1. Subtract 1 from both sides: $2\sin \theta = -1$
2. Divide by 2: $\sin \theta = -\frac{1}{2}$
3. The reference angle for $\frac{1}{2}$ is $\frac{\pi}{6}$. Since sine is negative in Quadrants III and IV:
   • $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$
   • $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

The key habit: always isolate the trig function completely before trying to find angles.

Calculator Use in Trigonometric Solutions

When the equation doesn't involve standard unit circle values, you'll need a calculator.

1. Isolate the trig function algebraically.

2. Use the inverse function on your calculator to get one solution:

   • $\sin^{-1}$ returns values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
   • $\cos^{-1}$ returns values in $[0, \pi]$
   • $\tan^{-1}$ returns values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

3. Find additional solutions. The calculator gives you only one angle, but there's usually a second solution within $[0, 2\pi)$. Use reference angles and quadrant analysis to find it.

Example: If $\cos \theta = 0.3$, your calculator gives $\theta \approx 1.2661$. The second solution is $\theta = 2\pi - 1.2661 \approx 5.0171$, since cosine is also positive in Quadrant IV.

Make sure your calculator is set to the correct mode (radians vs. degrees) for the problem.

Quadratic-Form Trigonometric Equations

Some trig equations look like quadratics in disguise. The trick is to substitute a single variable for the trig function, solve the quadratic, then convert back.

Steps:

1. Recognize the quadratic form: $a\sin^2 \theta + b\sin \theta + c = 0$ (or the same pattern with cosine).
2. Substitute $u = \sin \theta$ (or $u = \cos \theta$) to get $au^2 + bu + c = 0$.
3. Solve for $u$ by factoring, using the quadratic formula, or completing the square.
4. Substitute back and solve for $\theta$. Each valid value of $u$ gives its own set of angle solutions.
5. Check validity. If a solution for $u$ falls outside $[-1, 1]$, discard it, since sine and cosine can never exceed that range.

Example: Solve $2\sin^2 \theta - 3\sin \theta - 2 = 0$.

• Let $u = \sin \theta$: $2u^2 - 3u - 2 = 0$
• Factor: $(2u + 1)(u - 2) = 0$, so $u = -\frac{1}{2}$ or $u = 2$
• Discard $u = 2$ (impossible for sine). Solve $\sin \theta = -\frac{1}{2}$ using basic techniques.

Using Trigonometric Identities in Equations

When an equation contains more than one trig function, identities let you rewrite everything in terms of a single function so you can solve it.

Identities you'll use most often:

• Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$, which rearranges to $\sin^2 \theta = 1 - \cos^2 \theta$ or $\cos^2 \theta = 1 - \sin^2 \theta$
• Double angle formulas: $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ (also written as $2\cos^2 \theta - 1$ or $1 - 2\sin^2 \theta$)
• Power-reducing (half-angle) formulas: $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$

When to use them: If you see both $\sin \theta$ and $\cos \theta$ in the same equation, try using the Pythagorean identity to convert to just one function. If you see $\sin^2 \theta$ alongside terms in $\cos \theta$, replace $\sin^2 \theta$ with $1 - \cos^2 \theta$ to create a quadratic in cosine.

Example: Solve $\sin^2 \theta = \cos \theta + 1$ on $[0, 2\pi)$.

1. Replace $\sin^2 \theta$ with $1 - \cos^2 \theta$: $1 - \cos^2 \theta = \cos \theta + 1$

2. Rearrange: $\cos^2 \theta + \cos \theta = 0$

3. Factor: $\cos \theta(\cos \theta + 1) = 0$

4. Solve: $\cos \theta = 0$ gives $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$; and $\cos \theta = -1$ gives $\theta = \pi$

Multiple Angle Trigonometric Equations

These equations involve expressions like $\sin 2\theta$, $\cos 3\theta$, or $\tan \frac{\theta}{2}$. The angle inside the function isn't just $\theta$.

Approach 1: Solve for the multiple angle directly.

For $\sin 2\theta = \frac{1}{2}$ on $[0, 2\pi)$:

1. Let $\alpha = 2\theta$. Now solve $\sin \alpha = \frac{1}{2}$.
2. Adjust the interval: since $\theta \in [0, 2\pi)$, you need $\alpha \in [0, 4\pi)$.
3. $\alpha = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$
4. Divide each by 2: $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$

The common mistake here is forgetting to expand the interval in step 2. A multiple of $n$ inside the argument means you'll typically find more solutions than usual.

Approach 2: Use identities to expand.

Sometimes you can expand using sum/difference or double angle formulas:

• $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
• $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

This converts the equation into one involving only $\sin \theta$ and $\cos \theta$, which you can then solve with the techniques above.

Right Triangle Trigonometry Applications

Application problems ask you to find a missing side or angle in a real-world scenario involving right triangles.

Steps:

1. Sketch the triangle and label the known sides, angles, and the unknown you're solving for.

2. Choose the trig ratio that connects the known and unknown quantities:

   • $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
   • $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
   • $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

3. Set up the equation and solve algebraically.

4. Check that your answer is reasonable (a side can't be longer than the hypotenuse, angles in a triangle sum to 180°).

Example: A right triangle has a hypotenuse of 10 and an angle of 30°. Find the side opposite the 30° angle.

$\sin 30° = \frac{\text{opposite}}{10}$

$0.5 = \frac{\text{opposite}}{10}$

$\text{opposite} = 5$

Reciprocal Functions and Periodicity

Reciprocal trig functions show up in some equations and are defined as:

• $\csc \theta = \frac{1}{\sin \theta}$
• $\sec \theta = \frac{1}{\cos \theta}$
• $\cot \theta = \frac{1}{\tan \theta}$

To solve an equation like $\csc \theta = 2$, rewrite it as $\sin \theta = \frac{1}{2}$ and solve normally.

Periodicity is why trig equations have infinitely many solutions. Each trig function repeats at a regular interval:

• Sine and cosine repeat every $2\pi$
• Tangent and cotangent repeat every $\pi$

When writing general solutions, you add multiples of the period ($2\pi n$ or $\pi n$) to capture all solutions. When restricted to a specific interval, just list the solutions that fit.

The unit circle is your best visual tool for this. It shows you exactly where each trig function is positive or negative and helps you identify all solutions within a given interval.