Question

Solve the equation: sin−1(6x)+sin−1(6√3x)=−π2

Answer 1

In the attachment I have answered this problem.

I hope this answer Will be easy to understand

Answer 2

Answer: $x=-\frac{1}{12}$

Step-by-step explanation:

Here,

$sin^{-1}(6x)+sin^{-1}(6\sqrt{3})=-\frac{\pi}{2}$

$sin^{-1}(6x)+[\frac{\pi}{2} - cos^{-1}(6\sqrt{3})x]= -\frac{\pi}{2}$

$sin^{-1}(6x)=-\frac{\pi}{2} - \frac{\pi}{2} + cos^{-1}(6\sqrt{3})x$

$sin^{-1}(6x)=-\pi + cos^{-1}(6\sqrt{3})x$

$sin^{-1}(6x)= -[\pi-cos^{-1}(6\sqrt{3})x]$

$sin^{-1}(6x)= -cos^{-1}(-6\sqrt{3})x$

$-sin^{-1}(6x)= cos^{-1}(-6\sqrt{3})x$

$sin^{-1}(-6x)= cos^{-1}(-6\sqrt{3})x=\alpha$

$\implies sin^{-1}(-6x)=\alpha$  and $cos^{-1}(-6\sqrt{3})x=\alpha$

$\implies -6x = sin\alpha$  and $-6\sqrt{3}x=cos\alpha$

⇒ $\frac{-6x}{-6\sqrt{3}x}= tan\alpha$

⇒ $\frac{1}{\sqrt{3}}=tan\alpha$

⇒ $tan^{-1}\frac{1}{\sqrt{3}} = \alpha\implies \frac{\pi}{6}=\alpha$

⇒ $\frac{\pi}{6}=sin^{-1}(-6x)$

⇒ $sin \frac{\pi}{6}=-6x$

⇒ $\frac{1}{2}=-6x$

⇒ $-\frac{1}{12}=x$

Which is the required solution.