How many solutions does this composite function equation have?

How many solutions does this composite function equation have?

Question

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

1. has $0$ solutions;
2. has $1$ solution;
3. has $2$ solutions;
4. holds for all values of $x$.

Solution

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.