Question

The equation $si{n}^{-1}\left(\frac{3x}{5}\right)+si{n}^{-1}\left(\frac{4x}{5}\right)=si{n}^{-1}x$ has

• A. 3 roots
• B. the roots whose sum is zero
• C. roots which form an A.P.
• D. no root

Answer 1

Answer: (A), (B), (C)

The correct options are
A 3 roots
B the roots whose sum is zero
C roots which form an A.P.

Given

${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$

${\sin }^{-1}(\left(\frac{3x}{5}\right)\sqrt{1-{\left(\frac{4x}{5}\right)}^2}+\left(\frac{4x}{5}\right)\sqrt{1-{\left(\frac{3x}{5}\right)}^2})={\sin }^{-1}x$

${\sin }^{-1}(\frac{9x^2}{25}+\frac{16x^2}{25})={\sin }^{-1}x$

${\sin }^{-1}\frac{25x^2}{25}={\sin }^{-1}x$

${\sin }^{-1}{x}^2={\sin }^{-1}x$

$⟹x^2=|x|$

$⟹x=-1,0,1$

Sumof roots is $0$

No .of roots $=3$

The roots are in AP