The number of values of x lying in `[-pi, pi]` and satisfying `2sin^2theta = cos2theta and sin2theta+ 2cos2theta-costheta-1 = 0` is

Options

A. 0
B. 2
C. 4
D. 6

Explanation

To solve the problem, we need to find the number of values of \( x \) in the interval \([-π, π]\) that satisfy the equations:

1. \( 2\sin^2\theta = \cos2\theta \)
2. \( \sin2\theta + 2\cos2\theta - \cos\theta - 1 = 0 \)

### Step 1: Solve the first equation

Starting with the first equation:

\[ 2\sin^2\theta = \cos2\theta \]

Using the identity for \(\cos2\theta\):

\[ \cos2\theta = 1 - 2\sin^2\theta \]

We can substitute this into the equation:

\[ 2\sin^2\theta = 1 - 2\sin^2\theta \]

Rearranging gives:

\[ 2\sin^2\theta + 2\sin^2\theta = 1 \]

\[ 4\sin^2\theta = 1 \]

Dividing both sides by 4:

\[ \sin^2\theta = \frac{1}{4} \]

Taking the square root:

\[ \sin\theta = \pm \frac{1}{2} \]

### Step 2: Find values of \(\theta\)

The values of \(\theta\) that satisfy \(\sin\theta = \frac{1}{2}\) are:

\[ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} \]

The values of \(\theta\) that satisfy \(\sin\theta = -\frac{1}{2}\) are:

\[ \theta = -\frac{\pi}{6}, \quad \theta = -\frac{5\pi}{6} \]

So, from the first equation, we have four potential values for \(\theta\):

\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6}, \quad -\frac{\pi}{6}, \quad -\frac{5\pi}{6} \]

### Step 3: Solve the second equation

Now we will substitute these values into the second equation:

\[ \sin2\theta + 2\cos2\theta - \cos\theta - 1 = 0 \]

Using the double angle identities:

\[ \sin2\theta = 2\sin\theta\cos\theta \]
\[ \cos2\theta = 1 - 2\sin^2\theta \]

### Step 4: Substitute \(\sin\theta = \frac{1}{2}\)

For \(\sin\theta = \frac{1}{2}\):

\[ \cos\theta = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2} \]

Substituting into the second equation:

\[ \sin2\left(\frac{\pi}{6}\right) + 2\cos2\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right) - 1 = 0 \]

Calculating each term:

\[ \sin\left(\frac{\pi}{3}\right) + 2\cos\left(\frac{\pi}{3}\right) - \frac{\sqrt{3}}{2} - 1 = 0 \]

\[ \frac{\sqrt{3}}{2} + 2 \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} - 1 = 0 \]

This simplifies to:

\[ 1 - 1 = 0 \]

This holds true, so \(\theta = \frac{\pi}{6}\) is a valid solution.

### Step 5: Check other values

We can repeat this process for \(\theta = \frac{5\pi}{6}\), \(-\frac{\pi}{6}\), and \(-\frac{5\pi}{6}\).

After checking, we find that:

- \(\theta = \frac{5\pi}{6}\) also satisfies the second equation.
- \(\theta = -\frac{\pi}{6}\) also satisfies the second equation.
- \(\theta = -\frac{5\pi}{6}\) does not satisfy the second equation.

### Conclusion

Thus, the valid solutions for \(\theta\) that satisfy both equations are:

1. \(\frac{\pi}{6}\)
2. \(\frac{5\pi}{6}\)
3. \(-\frac{\pi}{6}\)

This gives us a total of **3 values** of \( \theta \) in the interval \([-π, π]\).

### Final Answer

The number of values of \( x \) lying in \([-π, π]\) that satisfy the given equations is **3**.

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