Question

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

$4{\cos }^2\theta -3=0$

Short answer

The solution set is$\left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6}\frac{11\mathrm{\pi }}{6}\right\}$.

Step-by-step solution

Step 1. Given Information 

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

$4{\cos }^2\theta -3=0$

Step 2. Firstly solving the equation

Add 3 on both side

$$\begin{gathered} 4{\cos }^2\theta -3+3=0+3 \\ 4{\cos }^2\theta =3 \end{gathered}$$

Divide by 4 on both side

$$\begin{gathered} \frac{4}{4}{\cos }^2\theta =\frac{3}{4} \\ {\cos }^2\theta =\frac{3}{4} \end{gathered}$$

Taking square root on both side

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

$\cos \theta =\frac{\sqrt{3}}{2}\mathrm{and}\cos \theta =-\frac{\sqrt{3}}{2}$

Step 3. After solving the equation cosθ=32  and  cosθ=-32

The period of the cosine function is $2\mathrm{\pi }$. In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which

$\cos \theta =\frac{\sqrt{3}}{2}:\theta =\frac{\mathrm{\pi }}{6}\mathrm{and}\theta =\frac{11\mathrm{\pi }}{6}$

Step 4. The period of the cosine function cosθ=-32 is 2π.

The period of the cosine function is $2\mathrm{\pi }$. In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which role="math" localid="1646583425702" $\cos \theta =-\frac{\sqrt{3}}{2}:\theta =\frac{5\mathrm{\pi }}{6}\mathrm{and}\theta =\frac{7\mathrm{\pi }}{6}$

The solution set is$\left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6}\frac{11\mathrm{\pi }}{6}\right\}$.