Question

Find the set of value of $x$ satisfying the equality $\sin (x-\frac{\pi }{4})-\cos (x-\frac{\pi }{4})=1$ and the inequality $\frac{2\cos 7x}{\cos 3+\sin 3}>2^{\cos 2x}$

Answer 1

$\sin (x-\frac{\pi }{4})-\cos (x-\frac{\pi }{4})=1$

$\frac{\sin x}{\sqrt{2}}-\frac{1}{\sqrt{2}}\cos x-[\frac{\cos A}{\sqrt{2}}+\frac{\sin A}{\sqrt{2}}]=1$

$\frac{\sqrt{2}\sin x}{2\pi /7}=1$

$\sin x=\frac{1}{\sqrt{2}}\Rightarrow x={45}^o$

$\frac{2\cos 7x}{\cos 3+\sin 3}>2^{\cos 2x}\to \pi$ $T(LCM$ of $\frac{2\pi }{7},\pi )$

$\Rightarrow 2\pi$

$\sin x=\frac{\pm }{\sqrt{2}}x=\frac{\pi }{4},\frac{3\pi }{4}$

$\left(1\right)\frac{\pi }{4}\Rightarrow \frac{2\cos 7\pi /4}{\sqrt{2}(\frac{\cos 3}{\sqrt{2}}+\frac{\sin 3}{\sqrt{2}})}>2^{\cos \pi /2}$

$\frac{1}{\cos (3-\pi /4)}>1$

$\frac{1}{\cos (3-\pi /4)}<0>1$

$x=\pi /4$ is not solution

$\left(2\right)\frac{3\pi }{4}\Rightarrow \frac{2\cos \frac{21\pi }{4}}{\sqrt{2}\cos (3-\pi /4)}>2^{\cos 3\pi /2}$

$\frac{2(-1/\sqrt{2})}{\sqrt{(}-ve)}>1$

$\frac{-1}{|\cos (3-\pi /4)|}>1$ $x=\frac{3\pi }{4}$

$x=2n\pi +3\pi /4$