Question

There exists a value of $\theta$ between 0 and $2\pi$ that satisfies the equation ${\sin }^4\theta -2{\sin }^2\theta -1=0.$

• A. True
• B. False

Answer 1

Given that equation is ${\sin }^4\theta -2{\sin }^2\theta -1=0$.

Lets compare with, $a{x}^2+bx+c=0$

$a=1,b=-2,c=-1$ and $x={\sin }^2\theta$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow {\sin }^2\theta =\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1\left)\right(-1)}}{2(-1)}$

$\Rightarrow {\sin }^2\theta =\frac{2\pm \sqrt{4+4}}{-2}$
$\Rightarrow {\sin }^2\theta =\frac{2\pm \sqrt{8}}{2}\Rightarrow {\sin }^2\theta =1\pm \sqrt{2}$
But ${\sin }^2\theta$ cannot be negative.

Therefore,
${\sin }^2\theta =\sqrt{2}+1$
But as $-1\le \sin \theta \le 1,\therefore {\sin }^2\theta \ne \sqrt{2}+1$
Thus, there is no value of $\theta$ which satisfic the given equation.
Therefore, statement is false.

The correct option is B False