Question

In Problems 69–78, solve each equation on the interval $0\le \theta <2\mathrm{\pi }$.

$\cos \left(2\theta \right)+6{\sin }^2\theta =4$.

Short answer

The solution set for the equation$\cos \left(2\theta \right)+6{\sin }^2\theta =4$is,$\left\{\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.

Step-by-step solution

Step 1 Given equation is,

$\cos \left(2\theta \right)+6{\sin }^2\theta =4$.

Th double angle formula for $\cos \left(2\theta \right)$is,$\cos \left(2\theta \right)=1-2{\sin }^2\theta$.

Use the formula in the given equation.

$1-2{\sin }^2\theta +6{\sin }^2\theta =4$

Now simplify it.

$$\begin{gathered} -2{\sin }^2\theta +6{\sin }^2\theta =4-1 \\ 4{\sin }^2\theta =3 \\ {\sin }^2\theta =\frac{3}{4} \end{gathered}$$

Step 2 Now take the square root on both the sides of the equation.

$$\begin{gathered} \sin \theta =\pm \sqrt{\frac{3}{4}} \\ \sin \theta =\pm \frac{\sqrt{3}}{2} \end{gathered}$$

Solving each equation in the interval $[0,2\mathrm{\pi })$we obtain $\theta =\frac{\mathrm{\pi }}{3},\theta =\frac{2\mathrm{\pi }}{3},\theta =\frac{4\mathrm{\pi }}{3}$and $\theta =\frac{5\mathrm{\pi }}{3}$.

Therefore the solution set is, $\left\{\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.