## vishweshshrimali5 one year ago Solve: $\large{\theta = \tan^{-1}(2\tan ^2\theta) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}$

1. vishweshshrimali5

@hartnn @amistre64 @Callisto @Hero

2. amistre64

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

3. vishweshshrimali5

Solve for theta

4. vishweshshrimali5

(general solutions)

5. amistre64

im just too sick to be able to concentrate on most of that :/

6. vishweshshrimali5

Its OK

7. vishweshshrimali5

8. amistre64

${\theta = \tan^{-1}(2\tan ^2\theta) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2 \theta }{ 5 + 4 \cos 2 \theta }}$ ${t = \tan^{-1}(2sec^2t-2) - \frac{1}{2} \sin^{-1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}$ ${2t = 2\tan^{-1}(2sec^2t-2) - \sin^{-1} \frac{3 \sin 2t }{ 5 + 4 \cos 2t }}$ yeah, im not going to be able to work it out in this condition :) good luck tho

9. hartnn

maybe use $$arcsin x= 2arctan (x/(1+\sqrt{1-x^2}))$$

10. harsh314

$\frac{ 1 }{ 2 }\sin^{-1} \frac{ 3\sin \theta }{ 5+4 \cos \theta }$be assumed as $\alpha$ now$\sin 2\alpha =\frac{ 3 \sin 2 \theta }{ 5+4\cos2 \theta }=m$ suppose now$m=\frac{ 6 \tan \theta }{ 9+\tan ^{2}\theta } ,$

11. harsh314

$\sqrt{\frac{ 1-m }{ 1+m }}=\tan{\frac{ \pi }{ 4 } -\alpha}$ $\sqrt{\frac{ 1-m }{ 1-m }}=\frac{ 3-\tan \theta }{ 3+\tan \theta }$ $\frac{ \pi }{ 4 }- \alpha=\tan^{-1} \sqrt{\frac{ 1-m }{ 1+m }}$ 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

12. hartnn

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

13. hartnn

to get other 2 roots, we can use synthetic division, taking x=1

14. harsh314

other values are tan theta =2,-1 ,thanks hartnn