Question

The number of values of x lying in $[-\pi, \pi]$ and satisfying $2\sin^2 \theta = \cos 2\theta$ and $\sin 2\theta + 2\cos 2\theta - \cos \theta - 1 = 0$ is

• A. 0
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (B)

2

Answer 2

Assuming the variable $\theta$ in the question is a typo and should be $x$:

The first equation is $2\sin^2 x = \cos 2x$. Using the identity $\cos 2x = 1 - 2\sin^2 x$, we get: $2\sin^2 x = 1 - 2\sin^2 x$ $4\sin^2 x = 1$ $\sin^2 x = \frac{1}{4}$ $\sin x = \pm \frac{1}{2}$

For $x \in [-\pi, \pi]$, the solutions are $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{\pi}{6}, -\frac{5\pi}{6}\}$.

For these values, $\cos 2x = 1 - 2\sin^2 x = 1 - 2(\frac{1}{4}) = \frac{1}{2}$.

The second equation is $\sin 2x + 2\cos 2x - \cos x - 1 = 0$. Substituting $\cos 2x = \frac{1}{2}$: $\sin 2x + 2(\frac{1}{2}) - \cos x - 1 = 0$ $\sin 2x + 1 - \cos x - 1 = 0$ $\sin 2x - \cos x = 0$

Using $\sin 2x = 2\sin x \cos x$: $2\sin x \cos x - \cos x = 0$ $\cos x (2\sin x - 1) = 0$

This means $\cos x = 0$ or $\sin x = \frac{1}{2}$.

We need to find the values from $\{\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{\pi}{6}, -\frac{5\pi}{6}\}$ that also satisfy $\cos x = 0$ or $\sin x = \frac{1}{2}$.

• For $x = \frac{\pi}{6}$: $\sin x = \frac{1}{2}$. This satisfies the condition.
• For $x = \frac{5\pi}{6}$: $\sin x = \frac{1}{2}$. This satisfies the condition.
• For $x = -\frac{\pi}{6}$: $\sin x = -\frac{1}{2}$ and $\cos x = \frac{\sqrt{3}}{2}$. This does not satisfy $\cos x = 0$ or $\sin x = \frac{1}{2}$.
• For $x = -\frac{5\pi}{6}$: $\sin x = -\frac{1}{2}$ and $\cos x = -\frac{\sqrt{3}}{2}$. This does not satisfy $\cos x = 0$ or $\sin x = \frac{1}{2}$.

Thus, the common solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. There are 2 such values.