DWG NO. 5.5 — Lesson 5 of 5

Solving Trigonometric Equations

Unit 5: Trigonometric Identities and Equations · ~25 min

Objective Solve trigonometric equations for all solutions in a given interval, or express the general solution using the function's period.

Everything up to this point has been about identities — statements true for every angle. A trig equation is the opposite: it's only true for specific angles, and the job is to find all of them. Because sine, cosine, and tangent repeat, a trig equation almost never has just one solution — it has a whole family. This lesson covers how to find every solution in a bounded interval like [0, 2π), and how to write the complete infinite family using the function's period.

The general approach

Isolate the trig function on one side, the same way you'd isolate x in an algebraic equation.
Find the reference angle using the inverse trig function, then use quadrant sign rules to locate every angle in [0, 2π) with that same trig value.
Add multiples of the period for a general solution: +2πk for sine and cosine, +πk for tangent and cotangent, since those repeat twice as often.
Watch for quadratic-looking equations in a single trig function (like 2sin²x − sinx − 1 = 0) — factor or use the quadratic formula treating sinx (or cosx, etc.) as the variable.

sinx = ½ has two solutions on the unit circle in [0, 2π): the horizontal line y = ½ crosses the circle at \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\)

Worked Example 1 · Basic equation on an interval

ProblemSolve 2sinx − 1 = 0 for x in [0, 2π).

1Isolate sinx: sinx = ½.

2Reference angle: \(\frac{\pi}{6.}\) Sine is positive in Quadrants I and II, giving x = \(\frac{\pi}{6}\) and x = π − \(\frac{\pi}{6}\) = \(\frac{5\pi}{6.}\)

x = \(\frac{\pi}{6,}\) \(\frac{5\pi}{6}\)

Worked Example 2 · Equation with a multiple angle

ProblemSolve cos2x = ½ for x in [0, 2π).

1Let u = 2x. Since x ranges over [0, 2π), u ranges over [0, 4π) — twice around the circle.

2Solve cosu = ½ over [0, 4π): u = \(\frac{\pi}{3,}\) \(\frac{5\pi}{3,}\) \(\frac{\pi}{3}\) + 2π = \(\frac{7\pi}{3,}\) \(\frac{5\pi}{3}\) + 2π = \(\frac{11\pi}{3.}\)

3Divide each by 2 to return to x: x = \(\frac{\pi}{6,}\) \(\frac{5\pi}{6,}\) \(\frac{7\pi}{6,}\) \(\frac{11\pi}{6.}\)

x = \(\frac{\pi}{6,}\) \(\frac{5\pi}{6,}\) \(\frac{7\pi}{6,}\) \(\frac{11\pi}{6}\)

Worked Example 3 · Quadratic-form equation, general solution

ProblemSolve 2sin²x − sinx − 1 = 0 for the general solution.

1This factors like a quadratic: (2sinx + 1)(sinx − 1) = 0.

2sinx = 1 gives x = \(\frac{\pi}{2}\) + 2πk.

3sinx = −½ has reference angle \(\frac{\pi}{6,}\) negative in Quadrants III and IV: x = π + \(\frac{\pi}{6}\) = \(\frac{7\pi}{6}\) + 2πk, or x = 2π − \(\frac{\pi}{6}\) = \(\frac{11\pi}{6}\) + 2πk.

x = \(\frac{\pi}{2}\) + 2πk, x = \(\frac{7\pi}{6}\) + 2πk, x = \(\frac{11\pi}{6}\) + 2πk

Guided practice

1.Solve 2cosx + 1 = 0 for x in [0, 2π).
Show answer

cosx = −½. Reference angle \(\frac{\pi}{3,}\) cosine negative in QII and QIII: x = \(\frac{2\pi}{3,}\) \(\frac{4\pi}{3}\).
2.Solve tanx = 1 for x in [0, 2π).
Show answer

Reference angle \(\frac{\pi}{4.}\) Tangent's period is π, so within [0,2π) there are two solutions: x = \(\frac{\pi}{4,}\) \(\frac{5\pi}{4}\).
3.Solve sin2x = 0 for x in [0, 2π).
Show answer

Let u = 2x ∈ [0,4π). sinu = 0 at u = 0, π, 2π, 3π. Dividing by 2: x = 0, \(\frac{\pi}{2,}\) π, \(\frac{3\pi}{2}\).
4.Find the general solution to cosx = \(\frac{\sqrt{2}}{2.}\)
Show answer

Reference angle \(\frac{\pi}{4,}\) cosine positive in QI and QIV: x = \(\frac{\pi}{4}\) + 2πk or x = \(\frac{7\pi}{4}\) + 2πk (equivalently \(\frac{-\pi}{4}\) + 2πk).
5.Explain why sinx = 2 has no solution.
Show answer

Sine is the y-coordinate on the unit circle, so its value is always between −1 and 1. Since 2 is outside that range, no angle can produce it — the equation has no real solution.

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