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

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.