Question

The value of $\theta$ satisfying the equation $(\cos \theta +\cos 3\theta -2\cos 2\theta )=0$

Answer 1

We have,

$\cos \theta +\cos 3\theta -2\cos 2\theta =0$

$\Rightarrow 2\cos 2\theta \cos \theta -2\cos 2\theta =0$...$(UsingFormulaofcosC+cosD)$

$\Rightarrow 2\cos 2\theta (\cos \theta -1)=0$

$\Rightarrow 2\cos 2\theta =0,\cos \theta -1=0$

$\Rightarrow \cos 2\theta =0,\cos \theta =1$

Now, $\cos 2\theta =0\Rightarrow 2\theta =(2n+1)\frac{\pi }{2},n\in Z$

$\Rightarrow \theta =(2n+1)\frac{\pi }{4},n\in Z$

And $\cos \theta =1$

$\cos \theta =\cos 0$

$\Rightarrow \theta =2m\pi \pm 0,m\in Z$

$\Rightarrow \theta =2m\pi m\in Z$

Hence $\theta =(2n+1)\frac{\pi }{4},2m\pi m,n\in Z$