<s>[INST] If $\sin^8\theta+\cos^8\theta=\frac{17}{32}$, find the value of $\theta$ using de Moivre's theorem. If
$$\sin^8\theta+\cos^8\theta=\frac{17}{32}$$
find the value of $\theta$ using  de Moivre's theorem. I tried a lot exapanding using Binomial theorem and taking real part and equating it given value. 
 [/INST] I try to give a solution with the given constraint.
Using de Moivre's formula we will show
$$
\begin{aligned}
\cos^4 t +\sin^4t &=\frac 1{4}\Big[\ \cos 4t+3\ \Big]\ ,
\\
\cos^8 t +\sin^8t &=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
\end{aligned}
$$
There is no prize of beauty we can win for the proof, so let us use $c:=\cos t$, $s=\sin t$ as shortcuts for an easy typing. The strategy is to write $\cos 4t$ and $\cos 8t$ in terms of an even polynomial in $c,s$, with no mixed monomials.
Let us get explicit relations:
$$
\begin{aligned}
\cos 4t
&=\text{Real part of }(\cos 4t +i\sin 4t)\\
&=\text{Real part of }(\cos t +i\sin t)^4\quad\text{(de Moivre)}\\
&=c^4-6c^2s^2+s^4\\
&=c^4-3(\underbrace{(c^2+s^2)^2}_{=1}-c^4-s^4)+s^4\\
&=4(c^4+s^4)-3\ ,\\
&\qquad\text{showing the first formula.}\\[2mm]
2c^2s^2
&=(c^2+s^2)^2-c^4-s^4=1-c^4-s^4\ ,\\
4c^4s^4
&=(1-c^4-s^4)^2\\
&=1+c^8+s^8-2c^4-2s^4+2c^4s^4\ ,\qquad\text{ so}\\
2c^4s^4
&=1+c^8+s^8-2c^4-2s^4\ ,\qquad\text{ so}\\[2mm]
\cos 8t
&=\text{Real part of }(\cos 8t +i\sin 8t)\\
&=\text{Real part of }(\cos 4t +i\sin 4t)^2\qquad\text{(de Moivre)}\\
&=\cos^24t-\sin^24t\\
&=2\cos^24t-1\qquad\text{(or going straightforward here...)}\\
&=2(4c^4+4s^4-3)^2
\\
&=\dots
\ .
\end{aligned}
$$
Some further lines are showing the claimed formula for $\cos 8t$. (The term in $c^4s^4$ can be split as above to have only monomials in $c$, or in $s$, or constant ones.)
From the given relation we have to solve the equation in $t$ (instead of $\theta$, which is too hard to type):
$$
\frac {17}{32}
=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
$$
Possibly using de Moivre, 
we write again
$\cos 8t = 2\cos^2 4t-1$ and obtain an equation of second degree in $u:=\cos 4t\in[-1,1]$,
$$
\frac {17}{32}
=\frac 1{64}\Big[\ 2u^2+28u +34\ \Big]\ .
$$
Now $17/32$ cancels on both sides, we get $2(u^2+17u)=0$, equivalently (for $|u|\le 1$) $u=0$, i.e. $4t$ is an odd multiple of $\pi/2$, i.e. $t$ is an odd multiple of $\pi/8$.
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