Question

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

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

Answer 1

Answer: (A)

We have, $si{n}^{-1}\frac{3x}{5}+si{n}^{-1}\frac{4x}{5}=si{n}^{-1}x$
$\Rightarrow si{n}^{-1}\left\{\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}\right\}=si{n}^{-1}x$
$\Rightarrow 3x\sqrt{25-16x^2}+4x\sqrt{25-9x^2}=25x$
$\Rightarrow x=0$ 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.