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Solve: \[\large{\theta = \tan^{-1}(2\tan ^2\theta) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\]

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and what is it we are looking to do with this monstrocity?
Solve for theta

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im just too sick to be able to concentrate on most of that :/
Its OK
Eventhough Thanks for your time
\[{\theta = \tan^{-1}(2\tan ^2\theta) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\] \[{t = \tan^{-1}(2sec^2t-2) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}\] \[{2t = 2\tan^{-1}(2sec^2t-2) - \sin^{-1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}\] yeah, im not going to be able to work it out in this condition :) good luck tho
maybe use \(arcsin x= 2arctan (x/(1+\sqrt{1-x^2}))\)
\[\frac{ 1 }{ 2 }\sin^{-1} \frac{ 3\sin \theta }{ 5+4 \cos \theta }\]be assumed as \[ \alpha\] now\[\sin 2\alpha =\frac{ 3 \sin 2 \theta }{ 5+4\cos2 \theta }=m\] suppose now\[m=\frac{ 6 \tan \theta }{ 9+\tan ^{2}\theta } ,\]
\[\sqrt{\frac{ 1-m }{ 1+m }}=\tan{\frac{ \pi }{ 4 } -\alpha}\] \[\sqrt{\frac{ 1-m }{ 1-m }}=\frac{ 3-\tan \theta }{ 3+\tan \theta }\] \[\frac{ \pi }{ 4 }- \alpha=\tan^{-1} \sqrt{\frac{ 1-m }{ 1+m }}\] substituting these values in the equation we obtain a cubic equation\[\tan ^{3}\theta -3\tan \theta+2=0\] i don't know how to solve it plz kindly tell me still by hit and trial i got pi/4
to solve, x^3-3x+2=0 i would first note that 1-3+2 =0 hence (x-1) must be the factor, x=1 must be the root tan theta =1 gives theta = pi/4
to get other 2 roots, we can use synthetic division, taking x=1
other values are tan theta =2,-1 ,thanks hartnn

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