<s>[INST] A strange trigonometric equation Today,in our class, we received a trigonometric equation
$$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$$
and the question was to find the general solution of this equation. My approach was, at First, trying to show that there were no solutions using inequalities, but I failed. So, my last method was, expanding RHS by binomial theorem, and canceling some terms out, which at last gives a quadratic in $\sin{x}\cos{x}$. But this way was too long. 
Can anyone suggest or give a simpler method? I firmly believe there's one trick in ques to make it easier, which I cannot solve.
 [/INST] We know that $X^5+Y^5=(X+Y)(X^4-X^3Y+X^2Y^2-XY^3+Y^4)$; if we set $X=\sin^2x$ and $Y=\cos^2x$, the left hand side becomes
$$
(\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)
$$
or
$$
\sin^8x+\cos^8x+\sin^4x\cos^4x-\sin^2x\cos^2x(\sin^4x+\cos^4x)
$$
The first three terms can be written
$$
(\sin^4x+\cos^4x)^2-\sin^4x\cos^4x
$$
and 
$$
\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x
$$
so we get
$$
(1-2\sin^2x\cos^2x)^2-\sin^4x\cos^4x-\sin^2x\cos^2x(1-2\sin^2x\cos^2x)
$$
which further simplifies into
$$
1-4\sin^2x\cos^2x+4\sin^4x\cos^4x-\sin^4x\cos^4x
-\sin^2x\cos^2x+2\sin^4x\cos^4x
$$
that is
$$
1-5\sin^2x\cos^2x+5\sin^4x\cos^4x
$$
or
$$
1-\frac{5}{4}\sin^2(2x)+\frac{5}{16}\sin^4(2x)
$$
Can you go on?
</s>