Trigonometric equation $3\sin^2(x)+\sin^2(3x)=3\sin x\cdot\sin^2(3x)$

Question

Number of real solution of the equation $\displaystyle \sin^2(x)+\frac{1}{3}\sin^2(3x)=\sin(x)\cdot\sin^2(3x)$ in $x\in[0,2\pi]$

Put $\sin^2(x)=a$ and $\sin^2(3x)=b$ then $\displaystyle a+\frac{b}{3}=ab^2$

$3a+b=3ab^2$ Help me how to solve

Answer 1

Write $s=\sin x$. Then $\sin 3x=3s-4s^4$. Your original equation is $$3s^2+(3s-4s^3)^2=3s(3s-4s^3)^2$$ which simplifies to $$s^2(-12+27s+24s^2-72s^3-16s^4+48s^5)=0.$$ Numerical investigation suggests the bracket is nonzero on $[-1,1]$ so that $s=0$, or $x=n\pi$ ($n$ integer) is the only solution.