Solving Trigonometric Equations

You have solved equations before. Lots of them. You know how to isolate a variable, how to work backward through operations, how to check your answer. When someone hands you $2x + 5 = 11$, you do not panic - you subtract 5, divide by 2, and confidently write $x = 3$.

Trigonometric equations are the same idea, just with a twist: instead of solving for $x$ directly, you are solving for an angle whose sine, cosine, or tangent equals some value. The equation $\sin x = \frac{1}{2}$ is asking: “What angle has a sine of one-half?” You already know at least one answer to this - it is $30°$ (or $\frac{\pi}{6}$ radians). But here is where things get interesting: that is not the only answer.

So, if solving trigonometric equations seems scary to you, then know that you are not alone. The periodic nature of trig functions means there are often infinitely many solutions, and knowing which ones to report (and how) takes some getting used to. But you need to realize something: you already know how to do this. You may not believe it, but, really, you do. You know how to solve equations. You know your trig values. Now you are just putting those skills together.

Core Concepts

Identities vs. Equations: What Is the Difference?

Before diving in, let us clarify an important distinction. You have likely seen trigonometric identities like $\sin^2 x + \cos^2 x = 1$. An identity is true for all values of the variable (where both sides are defined). It is a fundamental relationship that holds universally.

A conditional equation, on the other hand, is only true for certain specific values. The equation $\sin x = \frac{1}{2}$ is not true for every angle - it is only true for particular angles. Your job is to find those angles.

When you solve a trigonometric equation, you are finding the specific angles that make the equation true. When you prove a trigonometric identity, you are showing that two expressions are equivalent for all angles.

The Challenge of Multiple Solutions

Here is what makes trigonometric equations unique: the sine and cosine functions repeat every $2\pi$ radians (or $360°$). The tangent function repeats every $\pi$ radians (or $180°$). This periodicity means that if $x$ is a solution, then $x + 2\pi$, $x + 4\pi$, $x - 2\pi$, and infinitely many other values are also solutions.

For example, $\sin(30°) = \frac{1}{2}$, but so does $\sin(150°)$. And $\sin(30° + 360°) = \sin(390°) = \frac{1}{2}$ as well. We have infinitely many solutions.

This leads to two different types of answers you might be asked for:

Solutions in a restricted interval - Most commonly $[0, 2\pi)$ or $[0°, 360°)$. This gives you a finite list of answers within one complete cycle.

General solutions - All possible solutions, expressed using a formula with an integer $n$ to represent the infinite family of answers.

Solving Basic Equations: $\sin x = k$

When you have $\sin x = k$ where $-1 \leq k \leq 1$, the process goes like this:

Step 1: Find the reference angle. Use $\sin^{-1}(|k|)$ to get the reference angle - the acute angle whose sine is $|k|$.

Step 2: Determine which quadrants. Sine is positive in Quadrants I and II, negative in Quadrants III and IV.

Step 3: Find all solutions in one period. Based on the sign of $k$ and the quadrants, write out the angles.

Step 4: If asked for general solutions, add $+ 2\pi n$ (where $n$ is any integer) to account for all periods.

For sine, if $\sin x = k$ and the reference angle is $\alpha$:

• If $k > 0$: solutions in $[0, 2\pi)$ are $x = \alpha$ and $x = \pi - \alpha$
• If $k < 0$: solutions in $[0, 2\pi)$ are $x = \pi + \alpha$ and $x = 2\pi - \alpha$
• If $k = 0$: solutions are $x = 0$ and $x = \pi$
• If $k = 1$: solution is $x = \frac{\pi}{2}$
• If $k = -1$: solution is $x = \frac{3\pi}{2}$

Solving Basic Equations: $\cos x = k$

For cosine, the process is similar:

Step 1: Find the reference angle using $\cos^{-1}(|k|)$.

Step 2: Determine quadrants. Cosine is positive in Quadrants I and IV, negative in Quadrants II and III.

Step 3: Find solutions in one period.

Step 4: Add $+ 2\pi n$ for general solutions.

For cosine, if $\cos x = k$ and the reference angle is $\alpha$:

• If $k > 0$: solutions in $[0, 2\pi)$ are $x = \alpha$ and $x = 2\pi - \alpha$
• If $k < 0$: solutions in $[0, 2\pi)$ are $x = \pi - \alpha$ and $x = \pi + \alpha$
• If $k = 0$: solutions are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$
• If $k = 1$: solution is $x = 0$
• If $k = -1$: solution is $x = \pi$

Solving Basic Equations: $\tan x = k$

Tangent behaves differently because it has period $\pi$ (not $2\pi$) and can take any real value (not just $-1$ to $1$).

For $\tan x = k$:

Step 1: Find the reference angle using $\tan^{-1}(|k|)$.

Step 2: Determine quadrants. Tangent is positive in Quadrants I and III, negative in Quadrants II and IV.

Step 3: Find solutions in one period. Because tangent has period $\pi$, you only need one solution - the others are exactly $\pi$ apart.

For tangent, if $\tan x = k$ and the reference angle is $\alpha$:

• If $k > 0$: principal solution is $x = \alpha$; general solution is $x = \alpha + \pi n$
• If $k < 0$: principal solution is $x = \pi - \alpha$; general solution is $x = \pi - \alpha + \pi n$
• If $k = 0$: solutions are $x = \pi n$ (meaning $0, \pi, 2\pi$, etc.)

Notice that for tangent, the general solution uses $+ \pi n$ instead of $+ 2\pi n$ because of the shorter period.

Using Inverse Trigonometric Functions

Your calculator’s inverse trig functions ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) give you only one answer - the principal value. But equations usually have more solutions.

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

The calculator gives you a starting point. Your understanding of the unit circle and quadrants helps you find all the other solutions.

Equations Requiring Factoring

Sometimes a trigonometric equation is not in its simplest form. You might need to factor it first, just like you would factor a polynomial equation.

Consider $2\sin^2 x - \sin x = 0$. This factors as: $$\sin x(2\sin x - 1) = 0$$

By the zero product property (which works for trig expressions just like it works for polynomials), either:

• $\sin x = 0$, which gives $x = 0, \pi$ in $[0, 2\pi)$, or
• $2\sin x - 1 = 0$, meaning $\sin x = \frac{1}{2}$, which gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$

Combining these, the solutions in $[0, 2\pi)$ are $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$.

Quadratic-Type Trigonometric Equations

Some equations look like quadratics if you squint at them. The equation $2\cos^2 x - \cos x - 1 = 0$ has the same structure as $2u^2 - u - 1 = 0$ where $u = \cos x$.

The strategy:

1. Let $u$ equal the trig expression (like $\cos x$)
2. Solve the quadratic in $u$
3. Substitute back and solve for $x$
4. Check that all solutions are valid (especially important if you squared both sides earlier)

This technique works beautifully because factoring and the quadratic formula do not care whether the variable is $x$ or $\cos x$ - the algebra is identical.

Equations Requiring Identities

Sometimes you need to use a trigonometric identity to rewrite an equation before you can solve it. Common situations include:

• Pythagorean identities: Use $\sin^2 x + \cos^2 x = 1$ to convert between sine and cosine
• Double angle formulas: Useful when you have both $\sin 2x$ and $\sin x$ in the same equation
• Sum-to-product or product-to-sum formulas: Can simplify certain equations

The goal is usually to get the equation in terms of a single trig function so you can solve it using the techniques above.

For instance, if you have $\sin^2 x + \cos x + 1 = 0$, you can replace $\sin^2 x$ with $1 - \cos^2 x$: $$1 - \cos^2 x + \cos x + 1 = 0$$ $$-\cos^2 x + \cos x + 2 = 0$$ $$\cos^2 x - \cos x - 2 = 0$$

Now you have a quadratic in $\cos x$ that you can factor or use the quadratic formula on.

Multiple-Angle Equations

Equations like $\sin 2x = \frac{1}{2}$ or $\cos 3x = -1$ require an extra step. The argument of the trig function is not just $x$ but a multiple of $x$.

The approach:

1. Let $u$ equal the multiple angle (so $u = 2x$ or $u = 3x$)
2. Solve for $u$ in the appropriate interval
3. Solve for $x$ by dividing
4. Adjust your interval for $u$ based on the original interval for $x$

For example, if solving $\sin 2x = \frac{1}{2}$ for $x \in [0, 2\pi)$:

Since $x \in [0, 2\pi)$, we have $2x \in [0, 4\pi)$ (multiply the interval bounds by 2).

Solve $\sin u = \frac{1}{2}$ for $u \in [0, 4\pi)$: $$u = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{6} + 2\pi, \frac{5\pi}{6} + 2\pi = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$$

Now divide by 2 to get $x$: $$x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$$

Writing General Solutions

When asked for the general solution (all solutions, not just those in one period), you express your answer using an integer parameter $n$.

For sine and cosine equations (period $2\pi$): $$x = \text{(solution)} + 2\pi n \quad \text{where } n \text{ is any integer}$$

For tangent equations (period $\pi$): $$x = \text{(solution)} + \pi n \quad \text{where } n \text{ is any integer}$$

For example, the general solution to $\cos x = \frac{1}{2}$ is: $$x = \frac{\pi}{3} + 2\pi n \quad \text{or} \quad x = \frac{5\pi}{3} + 2\pi n$$

This can sometimes be written more compactly as $x = \pm\frac{\pi}{3} + 2\pi n$.

Notation and Terminology

Term	Meaning	Example
Conditional equation	An equation true only for certain values	$\sin x = \frac{1}{2}$ (true only when $x = \frac{\pi}{6}, \frac{5\pi}{6}$, etc.)
General solution	All solutions, expressed using $+ 2\pi n$ or $+ \pi n$	$x = \frac{\pi}{6} + 2\pi n$
Principal solution	Solutions within one period, typically $[0, 2\pi)$	$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$
Reference angle	The acute angle used to find solutions in other quadrants	For $\sin x = \frac{1}{2}$, the reference angle is $\frac{\pi}{6}$
Period	The length of one complete cycle	$2\pi$ for sine and cosine; $\pi$ for tangent

Examples

Key Properties and Rules

Strategy Summary by Equation Type

Equation Type	Strategy
$\sin x = k$	Find reference angle, use quadrant signs, add $2\pi n$ for general solution
$\cos x = k$	Find reference angle, use quadrant signs, add $2\pi n$ for general solution
$\tan x = k$	Find reference angle, use quadrant signs, add $\pi n$ for general solution
Quadratic in trig function	Substitute $u = \sin x$ (or $\cos x$, etc.), solve quadratic, substitute back
Multiple angle ($\sin 2x = k$)	Let $u = 2x$, adjust interval, solve for $u$, then divide to find $x$
Mixed functions ($\sin x = \cos x$)	Divide to get tangent, or use identities to rewrite in terms of one function

Where Trig Functions Are Positive

Quadrant	Sine	Cosine	Tangent
I	+	+	+
II	+	-	-
III	-	-	+
IV	-	+	-

Remember: All Students Take Calculus (All, Sine, Tangent, Cosine are positive in Quadrants I, II, III, IV respectively).

Reference Angle Formulas

If the reference angle is $\alpha$ (always positive and acute):

Quadrant	Angle in terms of reference angle
I	$\theta = \alpha$
II	$\theta = \pi - \alpha$
III	$\theta = \pi + \alpha$
IV	$\theta = 2\pi - \alpha$

Common Exact Values to Know

Angle	$\sin$	$\cos$	$\tan$
$0$	$0$	$1$	$0$
$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$\frac{\pi}{2}$	$1$	$0$	undefined

General Solution Patterns

For $\sin x = k$ with solutions $x = \alpha$ and $x = \beta$ in $[0, 2\pi)$:

• General solution: $x = \alpha + 2\pi n$ or $x = \beta + 2\pi n$

For $\cos x = k$ with solutions $x = \alpha$ and $x = 2\pi - \alpha$ in $[0, 2\pi)$:

• General solution: $x = \pm\alpha + 2\pi n$ (a compact form)

For $\tan x = k$ with solution $x = \alpha$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$:

• General solution: $x = \alpha + \pi n$

Real-World Applications

Finding Times of High and Low Tide

Tides are caused by the gravitational pull of the moon and sun on Earth’s oceans. The water level at a coastal location follows a roughly sinusoidal pattern:

$$h(t) = A\sin(Bt + C) + D$$

where $h$ is the water height, $t$ is time, $A$ is the amplitude (how much the tide varies), $B$ relates to the period, $C$ is the phase shift, and $D$ is the average water level.

To find when high tide occurs, you solve $h(t) = A + D$ (maximum height), which reduces to solving $\sin(Bt + C) = 1$. To find when the water reaches a specific level for safe boat passage, you solve for when $h(t)$ equals that level. This is exactly the kind of trigonometric equation we have been studying.

Determining Daylight Hours

The number of daylight hours throughout the year follows a sinusoidal pattern (in most locations). A model might look like:

$$D(t) = A\sin\left(\frac{2\pi}{365}(t - C)\right) + 12$$

where $D$ is hours of daylight and $t$ is the day of the year. Solving $D(t) = 14$ (for example) tells you on which days of the year you have exactly 14 hours of daylight. This involves solving a trigonometric equation.

Resonance in Physics

In physics, resonance occurs when a system is driven at its natural frequency. Many systems exhibit oscillatory behavior described by sine and cosine functions. Engineers need to solve trigonometric equations to determine:

• At what frequencies dangerous resonance might occur
• When oscillations reach maximum or minimum amplitude
• How to design systems that avoid destructive resonance

Sound and Light Wave Analysis

Both sound and light can be modeled as waves with sinusoidal components. Sound engineers working with speakers, audio equipment, or concert halls solve trigonometric equations to find:

• When two sound waves will constructively or destructively interfere
• At what positions along a string the amplitude of a standing wave is zero (nodes)
• How to tune instruments to precise frequencies

In optics, trigonometric equations help determine where light waves will create bright or dark bands in interference patterns, which is essential for designing optical instruments and understanding phenomena like thin-film interference (the colorful swirls you see in soap bubbles).

Self-Test Problems

Summary

• Trigonometric equations are solved to find the angles that make them true. Unlike identities (which are true for all angles), equations are only satisfied by specific values.

• Basic equations like $\sin x = k$, $\cos x = k$, and $\tan x = k$ are solved using reference angles and knowledge of which quadrants have positive or negative values.

• Solutions in a restricted interval (like $[0, 2\pi)$) give a finite list of answers. General solutions use $+ 2\pi n$ (for sine and cosine) or $+ \pi n$ (for tangent) to represent all infinitely many solutions.

• Quadratic-type equations like $2\cos^2 x - \cos x - 1 = 0$ are solved by substituting $u = \cos x$, solving the quadratic, then substituting back and solving for $x$.

• Multiple-angle equations like $\sin 2x = k$ require adjusting the interval (multiply by the coefficient of $x$), solving for the multiple angle, then dividing to get $x$.

• Equations with mixed functions can often be converted to a single function by dividing (to get tangent) or using identities.

• The inverse trig functions on your calculator give only the principal value. You must use your understanding of the unit circle to find all solutions.

• Real-world applications include modeling tides, daylight hours, wave interference, and any phenomenon that repeats periodically.

The key insight is this: solving trigonometric equations is not fundamentally different from solving other equations. You are still isolating, factoring, and using inverse operations. The new wrinkle is that trig functions repeat, so you often have multiple (even infinitely many) solutions. Once you internalize the pattern of finding reference angles and accounting for all quadrants, these problems become routine. And that is good news - because trigonometric equations appear everywhere in physics, engineering, and the sciences.