A community for students.
Here's the question you clicked on:
 0 viewing
vishweshshrimali5
 3 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
 3 years ago
Solve: \[\large{\theta = \tan^{1}(2\tan ^2\theta)  \frac{1}{2} \sin^{1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}\]

This Question is Closed

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

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

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

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

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

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

amistre64
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1maybe use \(arcsin x= 2arctan (x/(1+\sqrt{1x^2}))\)

harsh314
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1to get other 2 roots, we can use synthetic division, taking x=1

harsh314
 3 years ago
Best ResponseYou've already chosen the best response.1other values are tan theta =2,1 ,thanks hartnn
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.