Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

and what is it we are looking to do with this monstrocity?
Solve for theta

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

(general solutions)
im just too sick to be able to concentrate on most of that :/
Its OK
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question