Question

Which of the following is a solution to 4cos^2x-1=0? 60° 150° 210° 330°

Answer 1

Answer:

Option A- 60°            

Step-by-step explanation:

Given : Function - $4cos^2x-1=0$

To solve the function

Step 1 - Write the function

$4cos^2x-1=0$

Step 2- Add 1 both side

$4cos^2x=1$

Step 3- Divide 4 both side

$cos^2x=\frac{1}{4}$

Step 4 - Taking root both side

$cosx=\sqrt{\frac{1}{4}}$

Step 5 - solve

$cosx=\pm{\frac{1}{2}}$

Either $cosx={\frac{1}{2}}$ or $cosx=-{\frac{1}{2}}$

When $cosx={\frac{1}{2}}$

            $\rightarrow cosx={cos\frac{\pi}{3}}$

            $\rightarrow x={\frac{\pi}{3}} or 60^{\circ}$

or when  $cosx=-{\frac{1}{2}}$

               $\rightarrow cosx={cos(\pi-\frac{\pi}{3})}$

            $\rightarrow x={\frac{2\pi}{3}} or 120^{\circ}$

Therefore, Option A is correct solution of function is at 60°.

Answer 2

For have alook the photo