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

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The equation $(1+\sin^4 2\theta) = 17(1+\sin 2\theta)^4$ is analyzed for solutions in the interval $[0, 2\pi]$. By substituting $\sin 2\theta$ with $x$, the problem is transformed into finding roots of the polynomial $(x + 2)(2x + 1)(4x^2 + 7x + 4) = 0$. A graphical approach indicates a solution exists for $x$ in the range $(-1, 0)$. Further investigation reveals multiple corresponding values of $\theta$ for this range. The discussion seeks confirmation on the correctness of the solution method and the number of solutions found.
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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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