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Solve for x: $\sin^2{x}+(\sin^2{3x})/4=\sin{x}\sin^2{3x}$ I reached upto $\sin x =(1/ 2) \sin^2 {3x}$ and $\sin {6x}=0$.

After that?

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  • $\begingroup$ Let $a=\sin x$. Then $\sin 3x=3a-4a^3$ and: $$a^2+\frac{(3a-4a^3)^2}{4}=a(3a-4a^3)^2$$ $a=0$ is a solution, i.e. $x=\pi k,\, k\in\Bbb Z$. Let $a\neq 0$. Then: $$4+(3-4a^2)^2=4a(3-4a^2)^2$$ $$64 a^5 - 16 a^4 - 96 a^3 + 24 a^2 + 36 a - 13 = 0$$ $$(2 a-1)^2 (16 a^3+12 a^2-16 a-13) = 0$$ $a=\frac{1}{2}$ is a solution, i.e. $x=\frac{\pi}{6}+2\pi k$ or $x=\frac{5\pi}{6}+2\pi k$, where $k\in\Bbb Z$. You're left with solving $16 a^3+12 a^2-16 a-13=0$, which will give no solutions to your equation, but is problematic, because one root is extremely close to $1$, but slightly larger. $\endgroup$ – user236182 Nov 29 '15 at 23:14
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let $\sin^2 (x)=a$. it will become as follows on simplifying, we get $16a^3-8a^2+a=0$, we get $a=o;a=\frac{1}{4}$ which gives $\sin (x)=\frac{1}{2},\sin (x) =-\frac{1}{2}$; as $\sin(x)=0$

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  • $\begingroup$ how about $\LaTeX$? $\endgroup$ – BCLC Nov 30 '15 at 7:38

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