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vishweshshrimali5
 2 years ago
Solve:
\[\large{\theta = \tan^{1}(2\tan ^2\theta)  \frac{1}{2} \sin^{1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\]
vishweshshrimali5
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0@hartnn @amistre64 @Callisto @Hero

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

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

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

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

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

amistre64
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1maybe use \(arcsin x= 2arctan (x/(1+\sqrt{1x^2}))\)

harsh314
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1to get other 2 roots, we can use synthetic division, taking x=1

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