Question

If :$\cos 2\theta =(\sqrt{2}+1)\left(\cos \theta -\frac{1}{\sqrt{2}}\right)$ then find $\theta$.

Answer 1

given $\cos 2\theta =(\sqrt{2}+1)(\cos \theta -\frac{1}{\sqrt{2}})$

use $\cos 2\theta =2{\cos }^2\theta -1$

$2{\cos }^2\theta -1=(\sqrt{2}+1)(\cos \theta -\frac{1}{\sqrt{2}})$

substitute $x=\cos \theta$ & get a quadratic in $x$.

$2x^2-1=(\sqrt{2}+1)x-\frac{(\sqrt{2}+1)}{\sqrt{2}}$

$\Rightarrow 2x^2-(\sqrt{2}+1)x+\frac{1}{\sqrt{2}}=0$

$\Rightarrow x^2-(\frac{1}{\sqrt{2}}+\frac{1}{2})x+\frac{1}{2\sqrt{2}}=0$

$\Rightarrow (x-\frac{1}{\sqrt{2}})(x-\frac{1}{2})=0\Rightarrow x=\frac{1}{\sqrt{2}},\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}},\frac{1}{2}\Rightarrow \theta ={45}^o,{315}^o,{60}^o,{300}^o$

these are principal solutions of $\theta$.

General solution of $\theta =2n\pi +\frac{\pi }{4},$

$2n\pi +\frac{7\pi }{4},$

$2n\pi +\frac{\pi }{3},$

$2n\pi +\frac{5\pi }{3}$