Review question

# How many solutions does this equation involving $\sin^8x$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8136

## Solution

In the range $0 \leq x < 2\pi$ the equation $(3 + \cos x)^2 = 4 - 2\sin^8x$ has

1. $0$ solutions,

2. $1$ solution,

3. $2$ solutions,

4. $3$ solutions.

The quickest way to see this is to notice that $(3+\cos x)^2 \geq (3-1)^2 = 4$ and $4 - 2\sin^8x \leq 4.$ So there can only be solutions when these hold with equality, i.e. when $\cos x = -1$ and $\sin x = 0$. In the range $0 \leq x < 2\pi$ this only occurs at $x = \pi$, so the answer is (b).