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

The correct option is B False

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

$\therefore$ ${\sin }^2\theta =\frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2}$

But ${\sin }^2\theta =\sqrt{2}+1$

But as $-1\le \sin \theta \le 1,{\sin }^2\theta \ne \sqrt{2}+1$

Thus there is no value of $\theta$ which satisfies the given equation.

Therefore, statement is false.