Trigonometric Equation involving powers

Question

$$\sin^{2017}(x)+\sin^{2017}(7x)=2$$

I'm unable to solve this equation. Any hints for solving this type of problem?

Answer 1

First of all, we have $sinx \leqslant 1$ for all x.

Thus,

$$ sin^{2017}(x) + sin^{2017}(7x) \leqslant 2$$

Consider that $sin(x) = 1$. We can use the comment made by @robjohn. Since $7 \equiv 3 (mod4)$, then $sin(7x) = -1$.

Hence,

$$ sin^{2017}(x) + sin^{2017}(7x) = 0$$

Thus, we have that $sin(x) < 1$. But in this case $sin^{2017}(x) < 1$. Therefore,

$$ sin^{2017}(x) + sin^{2017}(7x) < 2$$

we conclude that there are no solutions in $\mathbb{R}$.