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

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vishweshshrimali5Best ResponseYou've already chosen the best response.0
@hartnn @amistre64 @Callisto @Hero
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
and what is it we are looking to do with this monstrocity?
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
Solve for theta
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
(general solutions)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
im just too sick to be able to concentrate on most of that :/
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
Eventhough Thanks for your time
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[{\theta = \tan^{1}(2\tan ^2\theta)  \frac{1}{2} \sin^{1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\] \[{t = \tan^{1}(2sec^2t2)  \frac{1}{2} \sin^{1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}\] \[{2t = 2\tan^{1}(2sec^2t2)  \sin^{1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}\] yeah, im not going to be able to work it out in this condition :) good luck tho
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
maybe use \(arcsin x= 2arctan (x/(1+\sqrt{1x^2}))\)
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
\[\frac{ 1 }{ 2 }\sin^{1} \frac{ 3\sin \theta }{ 5+4 \cos \theta }\]be assumed as \[ \alpha\] now\[\sin 2\alpha =\frac{ 3 \sin 2 \theta }{ 5+4\cos2 \theta }=m\] suppose now\[m=\frac{ 6 \tan \theta }{ 9+\tan ^{2}\theta } ,\]
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
\[\sqrt{\frac{ 1m }{ 1+m }}=\tan{\frac{ \pi }{ 4 } \alpha}\] \[\sqrt{\frac{ 1m }{ 1m }}=\frac{ 3\tan \theta }{ 3+\tan \theta }\] \[\frac{ \pi }{ 4 } \alpha=\tan^{1} \sqrt{\frac{ 1m }{ 1+m }}\] 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
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
to solve, x^33x+2=0 i would first note that 13+2 =0 hence (x1) must be the factor, x=1 must be the root tan theta =1 gives theta = pi/4
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
to get other 2 roots, we can use synthetic division, taking x=1
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
other values are tan theta =2,1 ,thanks hartnn
 one year ago
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