Number of solutions to a given equation

Question

Any hints on how to proceed further?

Answer 1

Put $\;t=\sin x\;$, then the equation becomes

$$2^{t^2}+5\cdot 2^{1-t^2}=7\iff2^{2t^2}-7\cdot 2^{t^2}+10=0$$

Put now $\;y:=2^{t^2}\;$ and get the quadratic

$$y^2-7y+10=0\implies (y-5)(y-2)=0\implies\begin{cases}5=y_1=2^{t^2}\implies t^2=\frac{\log5}{\log2}...etc.\\{}\\ 2=y_2=2^{t^2}\implies t^2=1...etc.\end{cases}$$

Observe that $\;t=\sin x\;$ and thus only one of the above two options for $\;t^2\;$ is possible...end now the exercise.

Answer 2

Hint: try with $$\sin^2x+\cos^2x =1$$

Say $t= \sin^2x $ and then $\cos^2x =1-t$...