Solving Trigonometric Equations

If you have ever set a thermostat to a specific temperature and waited for the room to reach that level, you have some intuition for what we are about to do. Solving a trigonometric equation is like asking: “At what angle does the sine (or cosine, or tangent) reach this particular value?” The twist, and it is a significant one, is that trigonometric functions are periodic. They repeat their values over and over as the angle increases. This means that unlike most algebraic equations you have solved before, trigonometric equations typically have infinitely many solutions.

Do not let that intimidate you. The periodicity that creates infinite solutions also gives us a beautiful structure to work with. Once you find the solutions in one period, you can describe all the others with a simple pattern. By the end of this chapter, you will be confidently solving trigonometric equations of all kinds, from the basic to the genuinely challenging.

Core Concepts

The Nature of Trigonometric Equations

An algebraic equation like $x^2 = 4$ has a finite number of solutions (in this case, $x = 2$ or $x = -2$). But consider the trigonometric equation $\sin\theta = \frac{1}{2}$. If you think about the unit circle, sine equals $\frac{1}{2}$ at $\theta = 30°$ (or $\frac{\pi}{6}$). But sine also equals $\frac{1}{2}$ at $\theta = 150°$ (or $\frac{5\pi}{6}$). And because sine repeats every $360°$ (or $2\pi$), both of those angles plus any multiple of $360°$ also work.

This gives us two distinct types of answers:

1. Solutions in a specific interval (like $[0°, 360°)$ or $[0, 2\pi)$): A finite list of angles
2. General solutions: All solutions, typically written using a variable like $n$ to represent any integer

Understanding when to provide which type is crucial. Many problems will specify the interval; when they do not, you should provide the general solution.

Basic Strategy for Solving Trig Equations

The overall approach mirrors solving algebraic equations:

1. Isolate the trigonometric function on one side of the equation
2. Find the reference angle where the function takes that value
3. Determine all angles in the specified interval (considering which quadrants give the correct sign)
4. Write the general solution if required (adding the appropriate multiples of the period)

Let us see how this works with the three primary functions.

Solving $\sin\theta = k$

When you have an equation like $\sin\theta = \frac{1}{2}$:

Step 1: Check if the equation has solutions. Since sine only produces values between $-1$ and $1$, if $|k| > 1$, there are no solutions.

Step 2: Find the reference angle $\alpha$ where $\sin\alpha = |k|$. You can use inverse sine: $\alpha = \sin^{-1}(|k|)$.

Step 3: Determine which quadrants contain solutions:

• If $k > 0$: Quadrants I and II (where sine is positive)
• If $k < 0$: Quadrants III and IV (where sine is negative)
• If $k = 0$: On the axes ($\theta = 0°, 180°$)

Step 4: Find the specific solutions:

• Quadrant I: $\theta = \alpha$
• Quadrant II: $\theta = 180° - \alpha$ or $\theta = \pi - \alpha$
• Quadrant III: $\theta = 180° + \alpha$ or $\theta = \pi + \alpha$
• Quadrant IV: $\theta = 360° - \alpha$ or $\theta = 2\pi - \alpha$

Step 5 (General solution): Since sine has period $2\pi$, the general solution is: $$\theta = \alpha + 2\pi n \quad \text{or} \quad \theta = (\pi - \alpha) + 2\pi n$$ where $n$ is any integer.

Solving $\cos\theta = k$

The process is similar, but the quadrants differ:

• If $k > 0$: Quadrants I and IV (where cosine is positive)
• If $k < 0$: Quadrants II and III (where cosine is negative)
• If $k = 0$: On the axes ($\theta = 90°, 270°$)

For a reference angle $\alpha = \cos^{-1}(|k|)$:

• Quadrant I: $\theta = \alpha$
• Quadrant II: $\theta = 180° - \alpha$ or $\theta = \pi - \alpha$
• Quadrant III: $\theta = 180° + \alpha$ or $\theta = \pi + \alpha$
• Quadrant IV: $\theta = 360° - \alpha$ or $\theta = 2\pi - \alpha$

General solution: Since cosine has period $2\pi$: $$\theta = \alpha + 2\pi n \quad \text{or} \quad \theta = -\alpha + 2\pi n$$ which is often written more compactly as $\theta = \pm\alpha + 2\pi n$.

Solving $\tan\theta = k$

Tangent is different in two important ways:

1. It has period $\pi$ (not $2\pi$)
2. It can take any real value (no restriction on $k$)

For $\tan\theta = k$:

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

Step 2: Determine which quadrants contain solutions:

• If $k > 0$: Quadrants I and III (where tangent is positive)
• If $k < 0$: Quadrants II and IV (where tangent is negative)
• If $k = 0$: $\theta = 0°, 180°$ (on the x-axis)

General solution: Because tangent has period $\pi$, you only need one solution and then add multiples of $\pi$: $$\theta = \tan^{-1}(k) + \pi n$$ where $n$ is any integer.

This is simpler than sine and cosine because the shorter period means one formula captures all solutions.

Equations with Multiple Angles

What if the equation involves $2\theta$ or $3\theta$ instead of just $\theta$? The key insight is to solve for the multiple angle first, then divide.

For an equation like $\sin(2\theta) = \frac{\sqrt{3}}{2}$:

Step 1: Let $u = 2\theta$ and solve $\sin u = \frac{\sqrt{3}}{2}$.

Step 2: Find all solutions for $u$ in the appropriate interval. If you want $\theta$ in $[0, 2\pi)$, then $u = 2\theta$ ranges from $0$ to $4\pi$.

Step 3: Divide by 2 to get $\theta$.

The critical point: when the angle is multiplied by 2, you need twice as many solutions in the original interval. When multiplied by 3, three times as many. This is because squeezing the function horizontally creates more repetitions in the same interval.

Equations Requiring Identities

Some equations cannot be solved directly because they involve multiple trigonometric functions. The solution is to use identities to rewrite everything in terms of a single function.

Common scenarios:

• Pythagorean identity: If you have both $\sin\theta$ and $\cos\theta$, use $\sin^2\theta + \cos^2\theta = 1$ to eliminate one.
• Double-angle formulas: Equations with $\sin(2\theta)$ or $\cos(2\theta)$ can often be rewritten using single-angle expressions.
• Tangent quotient identity: If you have sine divided by cosine, that is tangent.

Equations That Factor Like Quadratics

Many trigonometric equations, especially after applying identities, become quadratic in form. For example: $$2\cos^2\theta - 3\cos\theta + 1 = 0$$

This looks exactly like $2x^2 - 3x + 1 = 0$. You can solve it the same way:

1. Factor: $(2\cos\theta - 1)(\cos\theta - 1) = 0$
2. Set each factor to zero: $\cos\theta = \frac{1}{2}$ or $\cos\theta = 1$
3. Solve each simpler equation

Sometimes you will need to use the quadratic formula when factoring is not obvious.

Using Inverse Trig Functions

The inverse trigonometric functions are essential tools for finding reference angles:

• $\sin^{-1}(k)$ gives the angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ whose sine is $k$
• $\cos^{-1}(k)$ gives the angle in $[0, \pi]$ whose cosine is $k$
• $\tan^{-1}(k)$ gives the angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent is $k$

Remember: these functions give you only ONE angle. It is your job to find the other solutions using the symmetry of the unit circle.

Checking for Extraneous Solutions

Sometimes the algebraic process introduces solutions that do not actually work in the original equation. This happens especially when you:

• Square both sides of an equation
• Multiply or divide by a trigonometric expression that might be zero

Always verify your solutions by substituting them back into the original equation.

Notation and Terminology

Term	Symbol/Expression	Meaning
Reference angle	$\alpha$	The acute angle formed with the x-axis
General solution	$\theta = \alpha + 2\pi n$	All solutions, using integer $n$
Principal solution	$\theta$ in $[0, 2\pi)$	Solutions in one complete period
Inverse sine	$\sin^{-1}(k)$ or $\arcsin(k)$	The angle whose sine is $k$
Inverse cosine	$\cos^{-1}(k)$ or $\arccos(k)$	The angle whose cosine is $k$
Inverse tangent	$\tan^{-1}(k)$ or $\arctan(k)$	The angle whose tangent is $k$
Period	$T$	The interval after which a function repeats
Extraneous solution	—	A “solution” that does not satisfy the original equation

Examples

Key Properties and Rules

Summary of Solution Patterns

For $\sin\theta = k$ (where $|k| \leq 1$):

• Find reference angle: $\alpha = \sin^{-1}(|k|)$
• If $k > 0$: Solutions in Quadrants I and II: $\theta = \alpha$ and $\theta = \pi - \alpha$
• If $k < 0$: Solutions in Quadrants III and IV: $\theta = \pi + \alpha$ and $\theta = 2\pi - \alpha$
• General solution: $\theta = \alpha + 2\pi n$ or $\theta = (\pi - \alpha) + 2\pi n$

For $\cos\theta = k$ (where $|k| \leq 1$):

• Find reference angle: $\alpha = \cos^{-1}(|k|)$
• If $k > 0$: Solutions in Quadrants I and IV: $\theta = \alpha$ and $\theta = 2\pi - \alpha$
• If $k < 0$: Solutions in Quadrants II and III: $\theta = \pi - \alpha$ and $\theta = \pi + \alpha$
• General solution: $\theta = \pm\alpha + 2\pi n$

For $\tan\theta = k$ (any real $k$):

• Find angle: $\alpha = \tan^{-1}(k)$
• General solution: $\theta = \alpha + \pi n$ (simpler because period is $\pi$)

Number of Solutions in $[0, 2\pi)$

Equation Type	Typical Number of Solutions
$\sin\theta = k$ (with $0 <	k
$\cos\theta = k$ (with $0 <	k
$\tan\theta = k$	2
$\sin(n\theta) = k$	Up to $2n$
$\cos(n\theta) = k$	Up to $2n$
Quadratic in trig function	Up to 4
$\sin\theta = \pm 1$ or $\cos\theta = \pm 1$	1
$\tan\theta = 0$	2

Common Pitfalls to Avoid

1. Forgetting that trig equations usually have multiple solutions. Never stop after finding just one angle.

2. Dividing by a trig function without checking if it could be zero. This loses solutions. Factor instead.

3. Not checking for extraneous solutions after squaring both sides or other non-reversible operations.

4. Confusing the range of inverse trig functions with the solution interval. The inverse function gives you one reference angle; you find the rest.

5. Forgetting to expand the range when solving for a multiple angle. If you want $\theta$ in $[0, 2\pi)$ and you have $2\theta$, look for solutions in $[0, 4\pi)$.

Useful Identities for Solving Equations

These identities often help convert complicated equations into solvable forms:

Pythagorean: $\sin^2\theta + \cos^2\theta = 1$

Double-angle (sine): $\sin(2\theta) = 2\sin\theta\cos\theta$

Double-angle (cosine): $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$

Tangent quotient: $\tan\theta = \frac{\sin\theta}{\cos\theta}$

Real-World Applications

Astronomy and Satellite Communication

Satellites in orbit pass in and out of communication range as they circle Earth. If a satellite’s elevation angle $\theta$ (the angle above the horizon) follows a sinusoidal pattern, engineers solve equations like $\sin\theta = 0.3$ to determine when the satellite rises above the minimum angle needed for communication. Each orbit produces two solutions (rising and setting), and over multiple orbits, the general solution describes all communication windows.

Sound and Music

When two musical notes are played together, the combined sound wave is described by the sum of two sine functions. Finding when this combined wave reaches a certain intensity involves solving trigonometric equations. The periodic nature of the solutions explains why musical patterns repeat in predictable ways.

Electrical Engineering

Alternating current (AC) voltage varies sinusoidally. Engineers solve equations like $V_0\sin(\omega t) = V_{\text{threshold}}$ to determine when a circuit triggers, when a safety cutoff activates, or when two signals are in phase. The general solution with period $\frac{2\pi}{\omega}$ describes every occurrence.

Mechanical Oscillations

A mass on a spring oscillates according to equations like $x(t) = A\cos(\omega t + \phi)$. Solving for when the mass passes through a particular position (like the equilibrium point) involves trigonometric equations. The multiple solutions correspond to the mass passing through that point repeatedly, once in each direction per oscillation.

Navigation and Surveying

Determining a ship’s position using celestial navigation requires solving spherical trigonometry equations. Even with modern GPS, surveyors solve trigonometric equations to determine angles and distances when mapping terrain. The existence of multiple solutions corresponds to geometric ambiguity that must be resolved using additional information.

Daylight Hours

The number of daylight hours varies sinusoidally throughout the year. Solving an equation like $D(t) = 12$ (finding when day and night are equal) yields the spring and fall equinoxes. Solving $D(t) = 14$ might tell you which days have exactly 14 hours of sunlight, useful for agricultural planning or solar energy calculations.

Self-Test Problems

Summary

• Trigonometric equations typically have infinitely many solutions because trigonometric functions are periodic. In a restricted interval like $[0, 2\pi)$, there are usually finitely many solutions.

• The basic approach is to isolate the trig function, find the reference angle, determine which quadrants contain solutions based on the sign, and then find all angles in those quadrants.

• General solutions use the period of the function. For sine and cosine, add $2\pi n$. For tangent, add $\pi n$.

• Multiple angle equations like $\sin(2\theta) = k$ require you to first solve for the multiple angle, then divide. Remember to expand your search interval accordingly.

• Many equations require identities to convert them into a solvable form. The Pythagorean identity and double-angle formulas are especially useful.

• Quadratic equations in trig functions are solved just like regular quadratics: factor or use the quadratic formula, then solve the simpler equations that result.

• Never divide by a trig expression that might equal zero. Factor instead, so you do not lose solutions.

• Always check your answers for extraneous solutions, especially after squaring both sides or performing other non-reversible operations.

• Inverse trig functions give you one angle; it is your responsibility to find all others using the symmetry of the unit circle and the periodicity of the functions.

• These skills appear throughout physics, engineering, and applied mathematics whenever periodic phenomena need to be analyzed. Mastering them now prepares you for more advanced work in calculus and beyond.