Question

The number of solutions of equation ${\sin }^4\theta -2{\sin }^2\theta +1=0$ which lie between $0$ and $2\pi$ is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

From the given equation, we have
${\sin }^2\theta =\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$
Now ${\sin }^2\theta \ne 1+\sqrt{2}$
$\because 1+\sqrt{2}>1$
Also ${\sin }^2\theta \ne 1-\sqrt{2}$
$\therefore$ $1-\sqrt{2}<0$ and ${\sin }^2\theta$ cannot be negative
$\therefore$ no value of $\theta$ satisfy the equation, hence no root lies in 0 and $2\pi$.