# If cos^-1 6x/1+9x^2 =-pi/2 +2 tan^-1 3x , then find the values of x .

Answers (1)

Solution:  We have ,

$\cos^{-1}\frac{6x}{1+9x^{2}}=-\frac{\pi}{2}+2 \tan^{-1}3x$

$\Rightarrow$       $\frac{\pi}{2}-\sin^{-1}\frac{6x}{1+9x^{2}}=-\frac{\pi}{2}+2\tan^{-1}3x$

$\Rightarrow$      $\pi-2\tan^{-1}3x=\sin^{-1}\frac{2(3x)}{1+(3x)^{2}}$

We know that ,       $\sin^{-1}\frac{2x}{1+x^{2}}=\pi-2\tan^{-1}x \hspace{0.5cm}if \hspace{0.2cm}x> 1$

Therefore                $3x> 1\Rightarrow x> \frac{1}{3}$     ,   Thus $x\in(\frac{1}{3},\infty ).$

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