MHB Finding Solutions for $(1+sin^42\theta) = 17(1+sin2\theta)^4$ in $[0,2\pi ]$

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Number of solutions of $(1+sin^42\theta) = 17(1+sin2\theta)^4$ in $[0,2\pi ]$

I solved this by assuming $sin2\theta$ as $x$ and then I draw graph and found there was a solution for $x$ belonging to $(-1,0)$ then I drew graph of $sin2\theta$ and check for what values of $\theta$ it is lying in $(0,-1)$. But I got multiple values... Is this correct?? Help
 
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DaalChawal said:
Number of solutions of $(1+sin^42\theta) = 17(1+sin2\theta)^4$ in $[0,2\pi ]$

I solved this by assuming $sin2\theta$ as $x$ and then I draw graph and found there was a solution for $x$ belonging to $(-1,0)$ then I drew graph of $sin2\theta$ and check for what values of $\theta$ it is lying in $(0,-1)$. But I got multiple values... Is this correct?? Help
$sin2\theta=x$
$(x + 2) (2 x + 1) (4 x^2 + 7 x + 4) = 0$
 
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