Question

In Problems 11–34, solve each equation on the interval $0\le \theta \le 2\pi$.

$2{\sin }^2\theta -1=0$

Short answer

The solution set is$\left\{\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right\}$.

Step-by-step solution

Step 1. Given Information

In the given problem we have to solve each equation on the interval$0\le \theta \le 2\pi$.

$2{\sin }^2\theta -1=0$

Step 2. Firstly solving the equation 2sin2θ-1=0

Add 1 on both side

$$\begin{gathered} 2{\sin }^2\theta -1+1=0+1 \\ 2{\sin }^2\theta =1 \end{gathered}$$

Divide by 2 on both side

$$\begin{gathered} \frac{2}{2}{\sin }^2\theta =\frac{1}{2} \\ {\sin }^2\theta =\frac{1}{2} \end{gathered}$$

Taking square root on both side

$$\begin{gathered} \sin \theta =\pm \sqrt{\frac{1}{2}} \\ \sin \theta =\pm \frac{1}{\sqrt{2}} \end{gathered}$$

$\sin \theta =\frac{1}{\sqrt{2}}\mathrm{and}\sin \theta =-\frac{1}{\sqrt{2}}$

Step 3. After solving the equation  sinθ=12  and  sinθ=-12.

In the interval $[0,2\mathrm{\pi })$, there are two angles$\theta$ for which$\sin \theta =\frac{1}{\sqrt{2}}:\theta =\frac{\mathrm{\pi }}{4}\mathrm{and}\theta =\frac{3\mathrm{\pi }}{4}$

Step 4. The period of the cosine function sinθ=-12 is 2π.

In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which $\sin \theta =-\frac{1}{\sqrt{2}}:\theta =\frac{5\mathrm{\pi }}{4}\mathrm{and}\theta =\frac{7\mathrm{\pi }}{4}$

The solution set is$\left\{\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right\}$.