Question

Find the general solution of the trignometric equation $3^{(\frac{1}{2}+{\log }_3(\cos x+\sin x))}-2^{{\log }_2(\cos x+\sin x)}=\sqrt{2}$

Answer 1

$3^{(\frac{1}{2}+{\log }_3(\cos x+\sin x\left)\right)}-2^{lo{g}_2(\cos x+\sin x)}=\sqrt{2}$

$\Rightarrow \sqrt{3}(\cos x+\sin x)-(\cos x+\sin x)=\sqrt{2}$

$\Rightarrow (\cos x+\sin x)(\sqrt{3}-1)=\sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{3}-1}$

$\Rightarrow \sin (x+\frac{\pi }{4})=\frac{\sqrt{3}+1}{2}$

$\therefore x+\frac{\pi }{4}=n\pi +(-1{)}^n{\sin }^{-1}\left(\frac{\sqrt{3}+1}{2}\right)$

$\Rightarrow x=x\pi -\frac{\pi }{4}+(-1{)}^n{\sin }^{-1}\left(\frac{\sqrt{3}+1}{2}\right)$