Math · Precalculus · Worked example

Solve a trigonometric equation by factoring

Solve 2cos²x − cos x − 1 = 0 on [0, 2π).

2 {cos}^{2} x - cos x - 1 = 0

Treat it as a quadratic

With u = cos x the equation is 2u² − u − 1 = 0, which factors as (2u + 1)(u − 1) = 0.

\begin{array}{l} 2 {cos}^{2} x - cos x - 1 = 0 \\ (2 cos x + 1) (cos x - 1) = 0 \end{array}

Set each factor to zero

Either cos x = −1/2 or cos x = 1.

cos x = - \frac{1}{2} or cos x = 1

Solve each on the unit circle

cos x = −1/2 in quadrants II and III with reference angle π/3, so x = 2π/3 or 4π/3. In [0, 2π), cos x = 1 only at x = 0.

\begin{array}{l} cos \frac{2 π}{3} = - \frac{1}{2} \\ cos \frac{4 π}{3} = - \frac{1}{2} \\ cos 0 = 1 \end{array}

Result

x = 0, 2π/3 or 4π/3.

Your turn

Solve 2sin²x = sin x on [0, 2π).

Show the answer and explanation

x = 0, π/6, 5π/6 or π.

Move everything to one side and factor: sin x(2 sin x − 1) = 0. So sin x = 0, giving x = 0 or π, or sin x = 1/2, giving x = π/6 or 5π/6. Dividing both sides by sin x would lose 0 and π.

\begin{array}{l} 2 {sin}^{2} x = sin x \\ sin x (2 sin x - 1) = 0 \end{array}

Keep exploring

In Graph, y = 2cos²x − cos x − 1 meets the x-axis at 0, 2π/3 and 4π/3. At 0 it only touches the axis, because the factor cos x − 1 never changes sign.

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