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

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SUMMARY

The equation $(1+\sin^4 2\theta) = 17(1+\sin 2\theta)^4$ has been analyzed for solutions within the interval $[0, 2\pi]$. By substituting $\sin 2\theta$ with the variable $x$, the problem was transformed into finding roots of the polynomial $(x + 2)(2x + 1)(4x^2 + 7x + 4) = 0$. The analysis revealed that there is at least one solution for $x$ in the range $(-1, 0)$, leading to multiple corresponding values of $\theta$. This confirms the existence of multiple solutions for the original equation.

PREREQUISITES
  • Understanding of trigonometric identities, specifically $\sin 2\theta$.
  • Knowledge of polynomial equations and their roots.
  • Graphing techniques for visualizing functions and their intersections.
  • Familiarity with the interval notation and solution sets in trigonometric contexts.
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  • Study the properties of the sine function and its transformations.
  • Learn how to solve polynomial equations using graphical methods.
  • Explore the implications of multiple solutions in trigonometric equations.
  • Investigate the behavior of $\sin 2\theta$ over the interval $[0, 2\pi]$.
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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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