Math · Precalculus · Worked example

Solve a basic trigonometric equation

Solve 2 sin x − 1 = 0, first on [0, 2π) and then for all real x.

2 sin x - 1 = 0

Isolate the sine

Add 1 to both sides, then divide by 2.

\begin{array}{l} 2 sin x - 1 = 0 \\ 2 sin x = 1 \\ sin x = \frac{1}{2} \end{array}

Sine is 1/2 at the reference angle π/6. Sine is positive in quadrants I and II, so the solutions in [0, 2π) are π/6 and π − π/6 = 5π/6.

x = \frac{π}{6} or x = \frac{5 π}{6}

Sine repeats every 2π, so adding any whole number of turns gives another solution. With k any integer:

x = \frac{π}{6} + 2 π k or x = \frac{5 π}{6} + 2 π k

\begin{array}{c} x = \frac{π}{6} + 2 π k \\ or x = \frac{5 π}{6} + 2 π k \end{array}

On [0, 2π): x = π/6 or x = 5π/6. For all real x: x = π/6 + 2πk or x = 5π/6 + 2πk, with k an integer.

Solve 2 cos x + √3 = 0 on [0, 2π).

Show the answer and explanation

cos x = −√3/2. The reference angle is π/6, and cosine is negative in quadrants II and III, so x = π − π/6 = 5π/6 or x = π + π/6 = 7π/6.

\begin{array}{l} 2 cos x + \sqrt{3} = 0 \\ cos x = - \frac{\sqrt{3}}{2} \end{array}

In Graph, y = sin x crosses the line y = 1/2 at x = π/6 ≈ 0.524 and 5π/6 ≈ 2.618, and again every 2π after that.

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