Question

The number of solutions of the equation $\sin \, 2x + 2 \, \sin \, x - \cos \, x - 1 = 0$ in the range $0 \leq x \leq 2 \pi$ is

• A. 3
• B. 4
• C. 2
• D. None of these

Answer 1

Answer: (A)

3

Answer 2

We have, $\sin\,2x + 2\sin\,x - \cos\,x - 1 = 0$
$\Rightarrow 2\sin\,x\, \cos\,x + 2\sin\,x - \cos\,x - 1 = 0$
$\Rightarrow 2\sin\,x(\cos\,x + 1) -1 (\cos\,x + 1) = 0$
$\Rightarrow (\cos\,x + 1)(2\sin\,x - 1) = 0$
$\Rightarrow \cos\,x + 1 = 0$ or $2\sin\,x - 1 = 0$
$\Rightarrow \cos\,x= -1$ or $\sin\,x=\frac{1}{2}$
$\Rightarrow \cos\, x= \cos\,\pi$
or $\sin\, x=\sin \frac{\pi}{6}, \sin \frac{5\pi}{6} \left[\therefore x \in\left[0, 2\pi\right]\right]$
$\Rightarrow x=\pi$ or $x=\frac{\pi}{6}, \frac{5\pi}{6}$
$\therefore x=\pi, \frac{\pi}{6}, \frac{5\pi}{6}$
Hence, number of solutions are $3$.