Question

Inverse trigonometry question

Answer 1

\[\large 3\sin^{-1} ( \frac{2x}{1+x^2})-4\cos^{-1}(\frac{1-x^2}{1+x^2})+2\tan^{-1}(\frac{2x}{1-x^2})=\frac{\pi}{3}\]
Solve for x.

Answer 2

I tried..substituting x=tan theta..I got

Answer 3

\[\LARGE 6 \theta-8 \theta+4 \theta=\frac{\pi}{3}\]
\[\LARGE 2 \theta=\frac{\pi}{3}=>2 \tan^{-1}x=\frac{\pi}{3}\]

Answer 4

\[\LARGE \frac{2x}{1-x^2}=\frac{\pi}{3}\]

Answer 5

It would be difficult to solve this equation now :( for x

Answer 6

Anyway,
\[\LARGE 6x=\pi-\pi x^2 => \pi x^2+6x-\pi=0\]

Answer 7

\[\LARGE -6 \pm \frac{\sqrt{36-4 \pi^2}}{2 \pi}\]
hmm

Answer 8

are there any real solutions?

Answer 9

actually the discriminant is 36+4pi^2