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 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
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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vishweshshrimali5
 one year ago
Best ResponseYou've already chosen the best response.0@hartnn @amistre64 @Callisto @Hero

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0and what is it we are looking to do with this monstrocity?

vishweshshrimali5
 one year ago
Best ResponseYou've already chosen the best response.0Solve for theta

vishweshshrimali5
 one year ago
Best ResponseYou've already chosen the best response.0(general solutions)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0im just too sick to be able to concentrate on most of that :/

vishweshshrimali5
 one year ago
Best ResponseYou've already chosen the best response.0Eventhough Thanks for your time

amistre64
 one year ago
Best 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

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

harsh314
 one year ago
Best 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\[m=\frac{ 6 \tan \theta }{ 9+\tan ^{2}\theta } ,\]

harsh314
 one year ago
Best 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1to 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

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

harsh314
 one year ago
Best ResponseYou've already chosen the best response.1other values are tan theta =2,1 ,thanks hartnn
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