Question

The number of solutions of cos4θ – 2cos2θ + 3sin6θ + 1 = 0 in [0, 2π] is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The correct answer is option (C): 3

Given, $cos2\theta=sin\theta$

$\Rightarrow 1-2sin^2\theta=sin\theta$

$\Rightarrow (2sin \theta-1)(sin\theta+1)=0$

$\Rightarrow sin\theta =\frac{1}{2},sin\theta=-1$

\Rightarrow \theta=\left \\{ \frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2} \right \\} as $\theta\in (0,2\pi)$

The number of solutions is 3.