# Solve the equation on the interval \parenthesis 0,2\pi \parenthesis. 12\cos^{2}x = 9

## Question:

Solve the equation on the interval {eq}~[0,2\pi) {/eq}.

{eq}12 \cos^{2}x = 9 {/eq}

## Solving Trigonometric Equations:

The substitution of the trigonometric ratio is applied for the trigonometric equation solution. If the equation is solved after the substitution, then the value of the trigonometric ratio will have to be found after the back substitution.

In the problem we have to solve the equation on the interval {eq}~[0,2\pi) {/eq}.

The equation is trigoneometric and is given as:

{eq}12 \cos^{2}x = 9 {/eq}

So if we take:

{eq}\cos \left(x\right)=u\\ {/eq}

Then we have the above equation as:

{eq}12u^2=9\\ \Rightarrow u=\frac{\sqrt{3}}{2},\:u=-\frac{\sqrt{3}}{2}\\ \Rightarrow \cos \left(x\right)=\frac{\sqrt{3}}{2},\:\cos \left(x\right)=-\frac{\sqrt{3}}{2}\\ {/eq}

Now when:

{eq}\cos \left(x\right)=-\frac{\sqrt{3}}{2}\\ {/eq}

we will have:

{eq}x=\frac{5\pi }{6},\:x=\frac{7\pi }{6}\\ {/eq}

and when :

{eq}\cos \left(x\right)=\frac{\sqrt{3}}{2}\\ {/eq}

we will have:

{eq}x=\frac{\pi }{6},\:x=\frac{11\pi }{6}\\ {/eq}

So the solution set in the restricted domain is:

{eq}x=\frac{\pi }{6},\:x=\frac{11\pi }{6},\:x=\frac{5\pi }{6},\:x=\frac{7\pi }{6}\\ {/eq} 