Review question

# How many solutions does this composite function equation have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9924

## Solution

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 $x$.

### 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.

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 $(2^{\sin^2(x)} -1)^2 = -1,$