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:

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

Rearranging terms:

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

Step 1: Identify Possible Roots Consider the quadratic equation in terms of \(\sin x\):

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

Let:

\[ y = \sin x. \]

The equation becomes:

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

Step 2: Apply Quadratic Formula Using the quadratic formula:

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

Step 3: Check Valid Solutions For \(y = \sin x\) to be a valid solution, we require:

\[ -1 \leq y \leq 1. \]

This constraint eliminates extraneous roots and restricts the possible values of \(x\) within the interval \(-\pi \leq x \leq \pi\).

Step 4: Evaluate Specific Cases - \(\sin x = -3\) (rejected, as \(\sin x\) must lie within \([-1, 1]\)). - \(\sin x = (x - 1)^2\).

Solving \(\sin x = (x - 1)^2\) within the interval \(-\pi \leq x \leq \pi\) yields two valid solutions.

Therefore, the number of solutions is 2.