Solve the equation: $ 3(\sin^3 x + \cos^3 x) + 4(\sin^7 x + \cos^7 x) = 7(\sin^5 x + \cos^5 x) $

Question

Problem

Solve the equation in $[0, \pi]$
$$
3(\sin^3 x + \cos^3 x) + 4(\sin^7 x + \cos^7 x)
= 7(\sin^5 x + \cos^5 x)
$$

• This problem is from the algebra round of a local high school competition that has ended.

I have not been able to get very far for this question, I can only see that we might try to use sum of cubes to simplify the first term into $3(\sin x + \cos x)(1 – \cos x \cdot \sin x)$. Could I receive some help for this question? Thank you.

Answer 1

$$ 3(\sin^3 x + \cos^3 x) + 4(\sin^7 x + \cos^7 x) = 7(\sin^5 x + \cos^5 x). $$

$$\begin{align} 3(\sin^3 x + \cos^3 x)-3(\sin^5 x + \cos^5 x) &= 4(\sin^5 x + \cos^5 x)- 4(\sin^7 x + \cos^7 x) \\ \\ 3(\sin^3 x \cos^2x+ \cos^3 x\sin^2x)&= 4(\sin^5 x \cos^2x+ \cos^5 x\sin^2x) \\ \\ 3\sin^2 x\cos^2 x(\sin x + \cos x)&= 4\sin^2 x\cos^2 x(\sin^3 x + \cos^3 x) \end{align}$$

so we get:

$$\sin x=0, ~~\cos x=0,~~\sin x + \cos x=0,~~3=4(\sin^2x+\cos^2x-\sin x \cos x)$$ which is equivalent to $$\sin x=0, ~~\cos x=0,~~\sin x + \cos x=0,~~\sin(2x)=\frac{1}2$$

Solution:

$$\boxed{x=0,~~x=\frac{\pi}{12},~~x=\frac{\pi}2,~~ x=\frac{3\pi}4,~~x=\frac{5\pi}{12},~~ x=\pi}$$