Here's the question you clicked on:
vishweshshrimali5
Solve: \[\large{\theta = \tan^{-1}(2\tan ^2\theta) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\]
@hartnn @amistre64 @Callisto @Hero
and what is it we are looking to do with this monstrocity?
Solve for theta
(general solutions)
im just too sick to be able to concentrate on most of that :/
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