Question

Number of values of $\theta$ which satisfy ${\sin }^4\theta -2{\sin }^2\theta -1=\theta$ is?

• A. $0$
• B. Infinite
• C. Equal to the number of values of $\theta$ satisfying $8{\sec }^2\theta -6\sec \theta +1=0$
• D. None of these

Answer 1

The correct option is A $0$

${\sin }^4\theta -2{\sin }^2\theta -1=0$

Let ${\sin }^2\theta =y\cdot$

$\therefore y^2-2y-1=0$

$y=\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$

${\sin }^2\theta =1+\sqrt{2}\text{or}{\sin }^2\theta =1-\sqrt{2}$

${\sin }^2\theta =1+\sqrt{2}$ will give value of

sin $\theta$ greater than

1 which is impossible

${\sin }^2\theta =1-\sqrt{2}$ is also impossible because ${\sin }^2\theta$ cannot be negative.

$\therefore$ there is no possible $\theta$ for above given trigmometric equation

. Answer: option: (A)