Question

Solve the following equation
$sin^{-1} \frac{3x}{5} + sin^{-1} \frac{4x}{5} = sin^{-1} \,x$

• A. $0,1, -1$
• B. $0, -1$
• C. $0,1$
• D. $1,-1$

Answer 1

Answer: (A)

We have, $sin^{-1} \frac{3x}{5} + sin^{-1} \frac{4x}{5} = sin^{-1} \,x$
$\Rightarrow sin^{-1}\left\{\frac{3x}{5}\sqrt{1-\frac{16x^{2}}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^{2}}{25}}\right\} = sin^{-1}\,x$
$\Rightarrow 3x\sqrt{25-16x^{2}}+ 4x\sqrt{25-9x^{2}} = 25x$
$\Rightarrow x = 0$ or, $3\sqrt{25-16x^{2}}+ 4\sqrt{25-9x^{2}} = 25$
$\Rightarrow 4\sqrt{25-9x^{2}} = 25 - 3\sqrt{25 - 16x^{2}}$
Squaring both sides, we get
$16\left(25 - 9x^{2}\right) = 625 + 9\left(25 - 16x^{2}\right)-150\sqrt{25-16x^{2}}$
$\Rightarrow 150\sqrt{25-16x^{2}} = 450$
$\Rightarrow \sqrt{25-16x^{2} } = 3$
Again squaring both sides, we get
$25 - 16x^{2} = 9$
$\Rightarrow x = \pm 1$
Hence, $x = 0$, $1$, $-1$ are the roots of the given equation.