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Unit 5 · Analytic Trigonometry
5.3 Solving
Trigonometric Equations
Objectives: Use standard algebraic techniques to solve trigonometric equations.
Solve equations of quadratic type. Solve equations involving multiple angles.
Isolate the trig function Factor — never divide u-substitution Pythagorean Identities Unit Circle Check for extraneous solutions
Launch
Vertical Board Sort

At your board: Sort the equations into four columns labeled 0 solutions, 1 solution, 2 solutions, and 4 solutions in the interval \(0 \le \theta < 2\pi\).

Do not solve each one completely yet. Use the unit circle, symmetry, and the range of each trig function to justify your choices.

0 solutions
1 solution
2 solutions
4 solutions
Card A
\[\sin\theta = \dfrac{1}{2}\]
Card B
\[\cos\theta = -1\]
Card C
\[\tan\theta = 0\]
Card D
\[\sin\theta = 2\]
Card E
\[\tan^2\theta = 1\]
Card F
\[\cos\theta = 0\]

Debrief prompt: Be ready to defend one placement your group thinks another group might disagree with.

Concept
Introduction

To solve a trigonometric equation, use standard algebraic techniques such as collecting like terms and factoring.

Your preliminary goal is to isolate the trigonometric function involved in the equation, then use the unit circle to identify all angles that satisfy it.

Restricted vs. General Solutions:
When a domain is given (e.g., \(0 \le \theta < 2\pi\)), list only the angles in that interval.
When no domain is given, express all solutions using \(+ 2\pi n\) (or \(+ \pi n\) for tangent/cotangent), where \(n\) is any integer.

Key reminder: After solving, always verify your answers by substituting back into the original equation — especially if you squared both sides, which can introduce extraneous solutions.

Caution
Never divide both sides by a trig function

⚠️ Do NOT divide both sides by a trig function to cancel it. You may be dividing by zero, and you will eliminate solutions.

For example: \(\quad 2\cos\beta\sin\beta = \sin\beta\)
Dividing by \(\sin\beta\) gives \(2\cos\beta = 1\) — but you lost the solution \(\sin\beta = 0\).
Correct approach: Move everything to one side and factor.

Exception — dividing to convert to another function:
You may divide when the goal is to change the form of an expression into another trig function, not to "cancel" a factor.

For example, dividing \(\sin x + \cos x = \tan x\) through by \(\sin x\) gives \(1 + \cot x = \sec x\) — a valid identity transformation, not solving.

Example 1
Solve where \(0 \le \theta < 2\pi\)
\[\tan^2\theta - 3 = 0\]
Strategy: Isolate tan²θ, take the square root, then use the unit circle.
\(\tan^2\theta - 3 = 0\)
\(\tan^2\theta = 3\) add 3 to both sides
\(\tan\theta = \pm\sqrt{3}\) take the square root of both sides — both \(+\) and \(-\)
\(\tan\theta = \sqrt{3} \;\Rightarrow\; \theta = \dfrac{\pi}{3},\, \dfrac{4\pi}{3}\) tangent is positive in QI and QIII
\(\tan\theta = -\sqrt{3} \;\Rightarrow\; \theta = \dfrac{2\pi}{3},\, \dfrac{5\pi}{3} \qquad \checkmark\) tangent is negative in QII and QIV — solutions: \(\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}\)
Example 2
Solve for all solutions
\[\cot x \cos^2 x = 2\cot x\]
Strategy: Move everything to one side, factor — do not divide by cot x.
\(\cot x \cos^2 x = 2\cot x\)
\(\cot x \cos^2 x - 2\cot x = 0\) move everything to one side
\(\cot x\,(\cos^2 x - 2) = 0\) factor out \(\cot x\) (do NOT divide — that eliminates solutions)
\(\cot x = 0 \quad \text{or} \quad \cos^2 x = 2\) \(\cos^2 x = 2 \Rightarrow \cos x = \pm\sqrt{2}\) — impossible since \(|\cos x| \le 1\)
\(x = \dfrac{\pi}{2} + \pi n, \quad n \in \mathbb{Z} \qquad \checkmark\) cotangent equals 0 wherever tangent is undefined: \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\)
Unit 5 · Section 5.3
Equations of
Quadratic Type
When a trig equation has the form \(a[\,f(x)\,]^2 + b[\,f(x)\,] + c = 0\),
substitute \(u = f(x)\) to turn it into a standard quadratic, factor, then back-substitute.
Example 3
Solve where \(0 \le \theta < 2\pi\)
\[2\cos^2\theta - 7\cos\theta + 3 = 0\]
Strategy: Substitute \(u = \cos\theta\) to form a standard quadratic, then factor.
\(2\cos^2\theta - 7\cos\theta + 3 = 0\)
\(\text{Let } u = \cos\theta: \quad 2u^2 - 7u + 3 = 0\) u-substitution turns this into a standard quadratic
\((2u - 1)(u - 3) = 0\) factor the quadratic
\(2\cos\theta - 1 = 0 \quad \Rightarrow \quad \cos\theta = \tfrac{1}{2}\) back-substitute \(u = \cos\theta\)
\(\cos\theta - 3 = 0 \quad \Rightarrow \quad \cos\theta = 3\) impossible — range of cosine is \([-1, 1]\), so no solution from this factor
\(\theta = \dfrac{\pi}{3},\; \dfrac{5\pi}{3} \qquad \checkmark\) \(\cos\theta = \tfrac{1}{2}\) in QI and QIV
Concept
Restricted domain vs. all solutions

The previous example asked for solutions only in \(0 \le \theta < 2\pi\) — one full cycle of the unit circle.

Sometimes you need all solutions over the entire domain of the function.

General solution notation:
For sine and cosine (period \(2\pi\)): add \(+\, 2\pi n\) to each solution.
For tangent and cotangent (period \(\pi\)): add \(+\, \pi n\) to each solution.
where \(n\) is any integer.

Example: \(\cos\theta = \tfrac{1}{2}\) has general solutions \(\theta = \dfrac{\pi}{3} + 2\pi n\) and \(\theta = \dfrac{5\pi}{3} + 2\pi n\).

Example 4
Solve for all solutions
\[2\sin^2 x + 3\cos x - 3 = 0\]
Strategy: Substitute a Pythagorean identity to write the equation in terms of one function, then factor.
\(2\sin^2 x + 3\cos x - 3 = 0\)
\(2(1 - \cos^2 x) + 3\cos x - 3 = 0\) substitute \(\sin^2 x = 1 - \cos^2 x\)
\(2\cos^2 x - 3\cos x + 1 = 0\) expand, collect like terms, and simplify
\((2\cos x - 1)(\cos x - 1) = 0\) factor: \(\cos x = \tfrac{1}{2}\) or \(\cos x = 1\)
\(x = \dfrac{\pi}{3} + 2\pi n,\quad x = \dfrac{5\pi}{3} + 2\pi n,\quad x = 2\pi n \qquad \checkmark\) \(\cos x = \tfrac{1}{2}\) gives QI and QIV; \(\cos x = 1\) gives \(x = 0\)
Example 5
Solve in \([0,\, 2\pi)\)
\[\cos x + 1 = \sin x\]
Strategy: Square both sides to eliminate the radical — but check for extraneous solutions!
\(\cos x + 1 = \sin x\)
\((\cos x + 1)^2 = \sin^2 x\) square both sides to get a common function
\(\cos^2 x + 2\cos x + 1 = 1 - \cos^2 x\) substitute \(\sin^2 x = 1 - \cos^2 x\)
\(2\cos^2 x + 2\cos x = 0 \;\Rightarrow\; 2\cos x(\cos x + 1) = 0\) collect and factor — giving \(\cos x = 0\) or \(\cos x = -1\)
\(\cos x = 0 \Rightarrow x = \tfrac{\pi}{2}, \tfrac{3\pi}{2} \qquad \cos x = -1 \Rightarrow x = \pi\) check \(x = \tfrac{3\pi}{2}\): \(\cos\tfrac{3\pi}{2} + 1 = 1\) but \(\sin\tfrac{3\pi}{2} = -1\) — extraneous!
\(x = \dfrac{\pi}{2},\quad x = \pi \qquad \checkmark\) squaring both sides can introduce extraneous solutions — always verify
Unit 5 · Section 5.3
Functions Involving
Multiple Angles
For equations of the form \(\sin(ku) = c\) or \(\cos(ku) = c\),
first solve for \(ku\), then divide by \(k\) to get \(u\).
Example 6
Solve in \([0,\, 2\pi)\)
\[3\tan\!\left(\dfrac{x}{2}\right) + 3 = 0\]
Strategy: Isolate the trig function, solve for \(x/2\), then multiply by 2. Discard any result outside \([0,2\pi)\).
\(3\tan\!\left(\dfrac{x}{2}\right) + 3 = 0\)
\(\tan\!\left(\dfrac{x}{2}\right) = -1\) subtract 3, divide by 3
\(\dfrac{x}{2} = \dfrac{3\pi}{4}\) reference angle \(\tfrac{\pi}{4}\); tangent is negative in QII — \(\tfrac{x}{2} = \tfrac{3\pi}{4}\) is the solution in \([0,\pi)\)
\(\dfrac{x}{2} = \dfrac{7\pi}{4} \;\Rightarrow\; x = \dfrac{7\pi}{2}\) the other possibility — but \(\dfrac{7\pi}{2} > 2\pi\), so it falls outside \([0, 2\pi)\)
\(x = \dfrac{3\pi}{2} \qquad \checkmark\) multiply the valid solution by 2: \(\dfrac{3\pi}{4} \times 2 = \dfrac{3\pi}{2}\)
Example 7
Solve where \(0 \le \theta < 2\pi\)  — calculator required
\[5\sin\theta\cos\theta + 4\cos\theta = 0\]
Strategy: Factor out \(\cos\theta\), then solve each factor. Use inverse sine for the non-special value.
\(5\sin\theta\cos\theta + 4\cos\theta = 0\)
\(\cos\theta\,(5\sin\theta + 4) = 0\) factor out \(\cos\theta\)
\(\cos\theta = 0 \;\Rightarrow\; \theta = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}\) cosine equals 0 at the top and bottom of the unit circle
\(5\sin\theta + 4 = 0 \;\Rightarrow\; \sin\theta = -\dfrac{4}{5}\) sine is negative — solutions are in QIII and QIV
\(\theta_{\text{ref}} = \sin^{-1}\!\left(\dfrac{4}{5}\right) \approx 0.927\) find the reference angle using the calculator (positive value)
\(\theta \approx \pi + 0.927 \approx 4.069, \quad \theta \approx 2\pi - 0.927 \approx 5.356 \qquad \checkmark\) QIII: \(\pi + \theta_{\text{ref}}\)   QIV: \(2\pi - \theta_{\text{ref}}\) — all four solutions: \(\frac{\pi}{2},\; \frac{3\pi}{2},\; 4.069,\; 5.356\)