Question

Solve for $x:si{n}^{-1}2x+si{n}^{-1}3x=\frac{\pi }{3}$

• A. $\sqrt{\frac{76}{3}}$
• B. $\sqrt{\frac{3}{76}}$
• C. $\frac{3}{\sqrt{76}}$
• D. $\frac{\sqrt{3}}{76}$

Answer 1

Answer: (B)

$si{n}^{-1}2x+si{n}^{-1}3x=\frac{\pi }{3}$
$\Rightarrow \frac{\pi }{2}-co{s}^{-1}2x+\frac{\pi }{2}-co{s}^{-1}3x=\frac{\pi }{3}$
$\Rightarrow co{s}^{-1}2x+co{s}^{-1}3x=\frac{2\pi }{3}$
$\Rightarrow co{s}^{-1}\left\{6x^2-\sqrt{1-{\left(2x\right)}^2}\sqrt{1-{\left(3x\right)}^2}\right\}=\frac{2\pi }{3}$
$\Rightarrow 6x^2-\sqrt{\left(1-13x^2+36x^4\right)}=-\frac{1}{2}$
$\Rightarrow {\left(6x^2+\frac{1}{2}\right)}^2=1-13x^2+36x^4$
$\Rightarrow 19x^2=\frac{3}{4}$
$\Rightarrow x=\pm \sqrt{\frac{3}{76}}$
But sum of two negative numbers cannot be $\frac{\pi }{3}$.
$\therefore x=\sqrt{\frac{3}{76}}$ is the only solution.