Question

Find $\theta$, if $0\le \theta \le {90}^o$ ,$\sqrt{2}\cos \theta =1$

Answer 1

The given trignometric function is $\sqrt{2}\cos \theta =1$ that is

$\cos \theta =\frac{1}{\sqrt{2}}........\left(1\right)$

We know that $\cos {45}^0=\frac{1}{\sqrt{2}}........\left(2\right)$

From equations 1 and 2 we get,

$\cos \theta =\cos {45}^0\Rightarrow \theta ={45}^0$

Hence, $\theta ={45}^0$.