Find the least positive angle satisfying the trigonometric equation

Question

$\sin^3 x+\sin^3 2x+\sin^3 3x=(\sin x+\sin 2x+\sin 3x)^3$. I did solve the question, but my method is highly tedious. I combined the sin and then opened the cubic.... Is there some trick? Something I am missing? Thanks.

Answer 1

Hint: Note this indenty $$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$$ so $$(\sin{x}+\sin{(2x)})(\sin{(2x)}+\sin{(3x)})(\sin{x}+\sin{(3x)})=0$$ then $$\sin{x}\cos{x}(2\cos{x}+1)(4\cos^2{x}+2\cos{x}-1)=0$$ then you can solve it

Answer 2

For $\sin ax+\sin bx=0$

Method $\#1:$

$\sin ax+\sin bx=0\iff\sin ax=-\sin bx=\sin(-bx)$ as $\sin(-A)=-\sin A$

$\implies ax=n\pi+(-1)^n(-bx)$ where $n$ is any integer

If $n$ is even $=2m$(say), $ax+2m\pi-bx\iff x=\dfrac{2m\pi}{a+b}$

Similarly for odd $n=2m+1$(say)

Method $\#2:$

Using Prosthaphaeresis Formula, $$\sin ax+\sin bx=2\sin\dfrac{(a+b)x}2\cos\dfrac{(a-b)x}2$$

If $\sin\dfrac{(a+b)x}2=0, \dfrac{(a+b)x}2=p\pi$

and if $\cos\dfrac{(a-b)x}2=0,\dfrac{(a-b)x}2=\dfrac{(2q+1)\pi}2$ where $p,q$ are integers