Question

Solve the following equation for x:

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$, then x is equal to
a) $0,\frac{1}{2}$
b) $1,\frac{1}{2}$
c) zero
d) $\frac{1}{2}$

Answer 1

Given

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\Rightarrow -2si{n}^{-1}x=\frac{\pi }{2}-si{n}^{-1}(1-x)\Rightarrow -2si{n}^{-1}x=co{s}^{-1}(1-x)[\because si{n}^{-1}(1-x)+co{s}^{-1}(1-x)=\frac{\pi }{2}]$
$\Rightarrow cos(-2si{n}^{-1}x)=1-x$ [multiplying both sides by cosx]
$\Rightarrow cos\left(2si{n}^{-1}x\right)=1-x[\because cos(-x)=+cosx]\Rightarrow [1-2si{n}^2(si{n}^{-1}x\left)\right]=1-x[\because cos2x=1-2si{n}^{2}x]\Rightarrow 1-2\left[sin\right(si{n}^{-1}x){]}^2=1-x[\because si{n}^{2}x=\left(sinx{)}^2\right]\Rightarrow 1-2x^2=1-x\Rightarrow 2x^2-x=0$
$\Rightarrow x(2x-1)=0\Rightarrow x=0$ or $2x-1=0\Rightarrow x=0or\frac{1}{2}$
But $x=\frac{1}{2}$ does not satisfy the given equation, so x = 0
Hence,the correct option is (c).

Alternate method.

Given, $si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$
Let $x=sin\theta \Rightarrow \theta =si{n}^{-1}x$, then
$si{n}^{-1}(1-sin\theta )-2\theta =\frac{\pi }{2}\Rightarrow si{n}^{-1}(1-sin\theta )=\frac{\pi }{2}+2\theta \Rightarrow 1-sin\theta =sin(\frac{\pi }{2}+2\theta )\Rightarrow 1-sin\theta =cos2\theta [\because sin(\frac{\pi }{2}+\theta )=cos\theta ]\Rightarrow 1-sin\theta =1-2si{n}^2\theta [\because cos2\theta =1-2si{n}^2\theta ]\Rightarrow 2si{n}^2\theta -sin\theta =0\Rightarrow sin\theta (2sin\theta -1)=0$
Either $sin\theta or2sin\theta -1=0$
$\Rightarrow x=0or2x-1=0[\therefore sin\theta =x]\Rightarrow x=0orx=\frac{1}{2}$
But $x=\frac{1}{2}$ does not satisfy the given equation. So, x=0
To check your answer
Put x=0 in the given equation,
$\therefore si{n}^{-1}(1-0)-2si{n}^{-1}0=\frac{\pi }{2}\Rightarrow \frac{\pi }{2}-2\times 0=\frac{\pi }{2}\Rightarrow \frac{\pi }{2}=\frac{\pi }{2}$
Put $x=\frac{1}{2}$ in the given equation,
$\therefore si{n}^{-1}(1-\frac{1}{2})-2si{n}^{-1}\frac{1}{2}=\frac{\pi }{2}\Rightarrow si{n}^{-2}\frac{1}{2}-2si{n}^{-1}\frac{1}{2}=\frac{\pi }{2}\Rightarrow \frac{\pi }{6}-2\times \frac{\pi }{6}=\frac{\pi }{2}\Rightarrow \frac{\pi -2\pi }{6}=\frac{\pi }{2}\Rightarrow \frac{-\pi }{6}\ne \frac{\pi }{2},sox=\frac{1}{2}$ is not possibile

Note: While solving the trigonometric equation, sometimes it may have same extra roots, so please take careful about it.