Question

The number of common solutions of the pair of equations \(2sin^2θ-2cos^2θ+1 = 0\) and \(2sin^2θ+3sinθ-2 = 0\) in the interval \([0,2π]\), is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Solve each equation on [0, 2 pi] and count the angles that satisfy both.

Answer 1

The Correct Option is B

Step 1: Work with sine only:
The first equation is $2\sin^2\theta - 2(1 - \sin^2\theta) + 1 = 0$, so $4\sin^2\theta = 1$, meaning $\sin\theta = \pm\frac12$.

Step 2: Second equation:
Gives $\sin\theta = \frac12$ only, since $\sin\theta = -2$ is out of range.

Step 3: Intersect:
Common solutions need $\sin\theta = \frac12$. In $[0, 2\pi]$ this holds at $\theta = \frac\pi6$ and $\frac{5\pi}6$. So there are 2.

Final Answer:
The number of common solutions is 2, option (B). \[ \boxed{2} \]