<s>[INST] Help in Solving a Trigonometric Equation Solve the equation $$\left(\sin x + \cos x\right)^{1+\sin(2x)} = 2$$ when $-\pi \le x \le \pi $ .
I have tried to use $\sin (2x) = 2\sin x \cos x$ identity but I this doesn't lead me to a conclusion. 
I will appreciate the help.
 [/INST] To solve $(\sin x+\cos x)^{1+\sin 2x}=2$ when $−\pi\leq x\leq \pi$.
Let $z=\sin x+\cos x$, then $z^2=1+\sin 2x$. Hence the equation is  equivalent to $z^{z^2}=2$ which has the only possible solution $z=\sqrt{2}$.
Hence $\sin x+\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})=\sqrt{2}$.
Thus $\sin(x+\frac{\pi}{4})=1$, which implies $x+\frac{\pi}{4}=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.
Or using   $z^2=1+\sin 2x$ with $z=\sqrt{2}$ gives  $1+\sin 2x=2$
Hence  $\sin 2x=1$ implies $2x=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.
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