$2^{\sin(x) + \cos(y)} = 1$ , $16^{\sin^2(x) + \cos^2(y)} = 4$ (system of equations)

Question

My progress so far:

$2^{\sin(x) + \cos(y)} = 1$

$16^{\sin^2(x) + \cos^2(y)} = 4$

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$\sin(x) + \cos(y) = 0$

$\sin^2(x) + \cos^2(y) = \frac12$

I think I'm on the right track, but I'm not sure how to continue. Thanks in advance to anyone who helps.

Answer 1

Suppose $\sin x + \cos y = 0$. Then $$\sin ^2x + \cos ^2y = (\sin x + \cos y) ^2 -2\sin x\cos y = -2\sin x\cos y. $$ Thus we have left to consider $$\begin{cases} \sin x + \cos y = 0 \\ \sin x\cos y = -\frac{1}{4} \end{cases} $$ Can you take it from here?

Answer 2

From the first equation you get $$\sin x = -\cos y$$ Substituting that in the second equation, you get, $$\dfrac12=\sin^2x+\cos^2y =2\sin^2x\\ \implies \cos2x = 1-2\sin^2x =\dfrac12=\cos\dfrac\pi3\\ \implies 2x = 2n\pi\pm\dfrac\pi3\\ \implies \boxed{x = n\pi\pm\dfrac\pi6}\hspace{1cm}n\in\mathbb Z$$ or you could have directly got that using the fact that $$\sin^2x=\sin^2\dfrac\pi6\implies x=n\pi\pm\dfrac\pi6$$

Similarly, for $y$, we have $$\cos^2y = \cos^2\dfrac\pi3\\ \implies \boxed{y = n\pi\pm\dfrac\pi3}\hspace{1cm}n\in\mathbb Z$$