In the range \(0\leq x < 2\pi\) the equation \[2^{\sin^2(x)} + 2^{\cos^2(x)} =2\]

has \(0\) solutions;

has \(1\) solution;

has \(2\) solutions;

holds for all values of \(x\).

### Method 1

In the range \(0\leq x < 2\pi\), we have \(0\leq \sin^2 x \leq 1\) and \(0\leq \cos^2 x \leq 1\). So \(1 \leq 2^{\sin^2(x)} \leq 2\) and \(1 \leq 2^{\cos^2(x)} \leq 2\).

This means that the only way the sum \(2^{\sin^2(x)} + 2^{\cos^2(x)}\) can be equal to \(2\) is if both \(2^{\cos^2(x)}\) and \(2^{\sin^2(x)}\) are equal to \(1\), which occurs exactly when \(\cos^2(x) = 0\) and \(\sin^2(x)=0.\)

But in the given range \(\cos^2(x) =0\) at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) whereas \(\sin^2(x)=0\) at \(x=0\) and \(x=\pi\) (in fact, \(\cos(x)\) and \(\sin(x)\) can *never* be zero together). So there is no common solution.

The answer is a.

### Method 2

Multiply the equation by \(2^{\sin^2(x)}\). This gives us

\[(2^{\sin^2(x)})^2 + 2^{\cos^2(x)+\sin^2(x)} =2(2^{\sin^2(x)})\] or \[(2^{\sin^2(x)})^2 -2(2^{\sin^2(x)}) + 2 = 0.\] We can rewrite this as \[(2^{\sin^2(x)} -1)^2 = -1,\]

which clearly has no real solutions, so the answer is a.