Chapter 7.3, Problem 24AYU

In Problems 13-36, solve each equation on the interval $0\le \theta <2\pi$.

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

To determine

To find: A linear trigonometric equation $4{\text{cos}}^2\theta -3=0$ for $\theta$.

Answer to Problem 24AYU

The solution set $=\left\{\pi /6,11\pi /6\right\}$ for $\text{cos}\theta =+\surd 3/2$ and $\left\{5\pi /6,7\pi /6\right\}$ for $\text{cos}\theta =-\surd 3/2$.

Explanation of Solution

Given:

A linear trigonometric equation $4{\text{cos}}^2\theta -3=0$.

Calculation:

To find $\theta$, we need to solve for $\text{cos}\theta$.

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

Add 3 on both sides.

$4{\text{cos}}^2\theta =3$

Divide 4 on both sides.

${\text{cos}}^2\theta =3/4$

Taking square root on both sides.

$\text{cos}\theta =\pm \surd 3/2$

In the interval $0\le \theta <2\pi$, or in the interval notation, $\left[0,2\pi \right)$, we need to find angles for which $\text{cos}\theta =\pm \surd 3/2$.

$\text{θ}=\pi /6,11\pi /6$ for $\text{cos}\theta =+\surd 3/2$ and $\theta =5\pi /6$, $7\pi /6$ for $\text{cos}\theta =-\surd 3/2$.

Therefore, the solution set $=\left\{\pi /6,11\pi /6,5\pi /6,7\pi /6\right\}$.