Question

The number of solutions of \[\sin^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0, \quad \text{where } -\pi \leq x \leq \pi,\] is

Answer 1

Correct Answer: 2

Given the equation:

\[ \sin^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0. \]

Rearranged equation:

\[ \sin^2 x - (x^2 - 2x - 2) \sin x - 3(x - 1)^2 = 0. \]

Step 1: Formulate Quadratic in \(\sin x\) Treat the equation as a quadratic in \(\sin x\):

\[ \sin^2 x - (x^2 - 2x - 2) \sin x - 3(x - 1)^2 = 0. \]

Let \(y = \sin x\). The equation transforms to:

\[ y^2 - (x^2 - 2x - 2)y - 3(x - 1)^2 = 0. \]

Step 2: Apply Quadratic Formula Applying the quadratic formula for \(y\):

\[ y = \frac{(x^2 - 2x - 2) \pm \sqrt{(x^2 - 2x - 2)^2 + 12(x - 1)^2}}{2}. \]

Step 3: Validate Solutions Valid solutions require \(-1 \leq y \leq 1\). This condition eliminates extraneous roots and restricts \(x\) to \(-\pi \leq x \leq \pi\).

Step 4: Analyze Potential Roots

• \(\sin x = -3\): Rejected because \(\sin x\) must be in \([-1, 1]\).
• \(\sin x = (x - 1)^2\): This equation has two valid solutions within \(-\pi \leq x \leq \pi\).

Consequently, there are 2 solutions.