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. 

HINT
According to the formula
\begin{align*}
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\end{align*}
We can rewrite the given expresion as
\begin{align*}
\sin^{8}(\theta) + \cos^{8}(\theta) & = [\sin^{4}(\theta) + \cos^{4}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [(\sin^{2}(\theta)+\cos^{2}(\theta))^{2} - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [1 - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = \left[1 - \frac{\sin^{2}(2\theta)}{2}\right]^{2} - \frac{\sin^{4}(2\theta)}{8} = 1 -\sin^{2}(2\theta) + \frac{\sin^{4}(2\theta)}{8} = \frac{17}{32}
\end{align*}
Therefore, if we make the substitution $y = \sin(2\theta)$, the given problem is equivalent to
\begin{align*}
& 4y^{4} - 32y^{2} + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2}) + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2} + 16) - 49 = 0 \Longleftrightarrow\\
& 4(y^{2}-4)^{2} = 49 \Longleftrightarrow y^{2} - 4 = \pm\frac{7}{2} \Longleftrightarrow y^{2} = 4\pm\frac{7}{2} \Longleftrightarrow y = \pm\frac{1}{\sqrt{2}} = \pm\sin\left(\frac{\pi}{4}\right)
\end{align*}
Can you proceed from here?
EDIT
According to the Newton's binomial theorem as well as the De Moivre's formula, we have
\begin{align*}
\cos(8\theta) + i\sin(8\theta) = [\cos(\theta) + i\sin(\theta)]^{8} = \sum_{k=0}^{8}\binom{8}{k}[\cos(\theta)]^{k}[i\sin(\theta)]^{8-k}
\end{align*}
As a consequence, you can obtain the value of the expression $\sin^{8}(\theta) + \cos^{8}(\theta)$ in terms of sines and cosines and solve the 
corresponding equation.
