Question

If $4{\cos }^2{x}^0-1=0$ and $0\le x^0\le {90}^0$, find $x$ in degrees.

Answer 1

The given equation is $4{\cos }^2{x}^o-1=0$

To find out,

The value of $x$ in degrees.

$\Rightarrow 4{\cos }^2{x}^o=1$

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

$\Rightarrow \cos x^o=\pm \frac{1}{2}$

Also, given that, $0\le x^o\le {90}^o$

Hence, $\cos x^o=\frac{1}{2}\left[\cos \theta \text{is positive in the first quadrant}\right]$

$\Rightarrow \cos x^o=\cos {60}^o[\because \cos {60}^o=\frac{1}{2}]$

$\therefore x={60}^o$

Hence, the value of $x$ is ${60}^o$.