NIMCET 2020 — Mathematics PYQ

Concept:

The solution to the quadratic equation $a{x}^2+bx+c=0$ is given by: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
- $\sin (-\theta )=-\sin \theta$.
- $\sin \theta =\sin (2n\pi +\theta )$.

Calculation:

${\sin }^2\mathbf{x}-\sin \mathbf{x}-2=0$
Using the formula for the roots of a quadratic equation.
$\Rightarrow \sin \mathbf{x}=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(-2)}}{2(1)}$
$\Rightarrow \sin \mathbf{x}=\frac{1+\sqrt{9}}{2}$ OR$\sin \mathbf{x}=\frac{1-\sqrt{9}}{2}$
$\Rightarrow \sin \mathbf{x}=2$ OR $\sin \mathbf{x}=-1.$
$\because -1\le \sin \mathbf{x}\le 1,\therefore \sin \mathbf{x}=2$is not possible
$\sin \mathbf{x}=-1=-\sin \frac{\pi }{2}=\sin \left(-\frac{\pi }{2}\right)=\sin \left(2\mathbf{n}\pi -\frac{\pi }{2}\right),\mathbf{n}\in \mathbf{Z}$
For n$=0,\mathbf{x}=-\frac{\pi }{2}.$ For n= 1, x= $\frac{3\pi }{2}.$ For n= 2, x= $\frac{7\pi }{2}.$ The only value of x on the interval 0\leq\mathbf{x}<2\pi is x$=\frac{3\pi }{2}$