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.
 A: $\displaystyle\cos2x=1-2\sin^2x=2\cos^2x-1$
Setting $\displaystyle\cos2x=u,$ we get $$\left(\frac{1-u}2\right)^5+\left(\frac{1+u}2\right)^5=\frac{29}{16}u^4$$
$$\iff2\left[1+\binom52u^2+\binom54u^4\right]=2^5\cdot\frac{29}{16}u^4$$
Again, $\displaystyle u^2=\frac{1+\cos4x}2$
A: $$ \sin^{10}x + \cos^{10}x = \frac{29}{16} \cos^4 2x $$
We can do some algebraic manipulation with the LHS in the following manner -
$$ \sin^{10}x + \cos^{10}x $$
$$\left(\frac{1- \cos 2x}{2}\right)^5 + \left(\frac{1+ \cos 2x}{2}\right)^5$$
$$ 2\times \frac{\left( \binom{5}{0} + \binom{5}{2} \cos^{2}2x + \binom{5}{4} \cos^4 2x \right)}{2^5}$$
Equate this to the RHS, and solve for the quadratic in $\cos^2 2x.$
A: 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?
A: can this approach be somehow helpful? set $sin^2x=u$ and $cos^2x=1-u$ similar to what is suggested in one of the comments above. Then, after some manipulations the equations simplify to $$-\frac{(48a^2 - 48a + 13)\cdot(8a^2 - 8a + 1)}{16}=0$$ (I used matlab's symbolic toolbox to obtain this) whose real solutions are $$u=\frac{\sqrt{2}}{4}+\frac{1}{2}$$ and $$u=\frac{1}{2}-\frac{\sqrt{2}}{4}$$
can you continue from here?
A: Rewrite $\cos(2x)=1-2\sin^2(x)$ and $\cos^{10}(x)=(1-\sin^2(x))^5$ and then factor
$\sin^{10}(x)+(1-\sin^2(x))^5-\frac{29}{16}(1-2\sin^2(x))^4$ to get
$$-\frac{1}{16}(48\sin(x)^4-48\sin(x)^2+13)(8\sin(x)^4-8\sin(x)^2+1)=0.$$ 
The first equation has no solutions since the discriminant is negative and the second has the solutions $\sin(x)=\pm\frac{1}{2}\sqrt{2-\sqrt{2}}$ giving the solutions $\frac{k\pi}{8}, k=\pm 1, \pm 3$.
