1. Given equation

We need to solve

$$\sin^2 x + (2+2x-x^2)\sin x - 3(x-1)^2 = 0,$$

for $-\pi \le x \le \pi$.

We are asked for the number of solutions.

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2. Treat it as a quadratic in $\sin x$

Let

$$y = \sin x.$$

Then the equation becomes

$$y^2 + (2+2x-x^2)y - 3(x-1)^2 = 0.$$

Now factor this quadratic.

We look for two expressions whose product is $-3(x-1)^2$ and whose sum is $2+2x-x^2$.

Observe that

$$3-x^2 = -(x^2-3).$$

Try

$$(y + 3 - x^2)(y + x - 1).$$

Expanding:

$$(y + 3 - x^2)(y + x - 1) = y^2 + \big((3-x^2)+(x-1)\big)y + (3-x^2)(x-1).$$

The coefficient of $y$ is

$$(3-x^2)+(x-1)=2+x-x^2,$$

which is not the required one.

So instead try

$$(y + x - 1)(y + 3 - x).$$

That gives constant term $(x-1)(3-x)$, so not correct.

Let us factor systematically. Since

$$-3(x-1)^2 = (x-1)(-3x+3),$$

try

$$(y- (x-1))(y + 3(x-1)).$$

Expanding:

$$(y-x+1)(y+3x-3) = y^2 + (2x-2)y -3(x-1)^2.$$

This gives coefficient $2x-2$, still not correct.

Now use the quadratic formula in $y$:

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

Let

$$A = 2+2x-x^2.$$

Then discriminant is

$$D = A^2 + 12(x-1)^2.$$

Compute it:

$$A = -x^2+2x+2.$$

So

$$A^2 = (-x^2+2x+2)^2 = x^4-4x^3+8x+4.$$

Also,

$$12(x-1)^2 = 12x^2-24x+12.$$

Hence

$$D = x^4-4x^3+12x^2-16x+16.$$

Now check if this factors:

$$x^4-4x^3+12x^2-16x+16 = (x^2-2x+4)^2.$$

Therefore,

$$\sqrt D = x^2-2x+4,$$

since $x^2-2x+4=(x-1)^2+3>0$.

So

$$y = \frac{-A \pm (x^2-2x+4)}{2}.$$

Since

$$-A = x^2-2x-2,$$

we get:

• With $+$ sign,

$$y = \frac{x^2-2x-2 + x^2-2x+4}{2} = x^2-2x+1 = (x-1)^2.$$

• With $-$ sign,

$$y = \frac{x^2-2x-2 - (x^2-2x+4)}{2} = -3.$$

Thus

$$\sin x = (x-1)^2 \quad \text{or} \quad \sin x = -3.$$

Since $\sin x\in[-1,1]$, the equation $\sin x=-3$ has no solution.

So we only need to solve

$$\sin x = (x-1)^2.$$

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3. Restrict possible values using ranges

Because $\sin x \in [-1,1]$ and $(x-1)^2 \ge 0$, we must have

$$0 \le (x-1)^2 \le 1.$$

So

$$|x-1| \le 1 \implies 0 \le x \le 2.$$

Thus any solution must lie in $[0,2]$.

Also on $[0,2]$, we have

$$\sin x > 0.$$

So solutions are possible there.

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4. Solve graphically / by sign analysis

Define

$$f(x)=\sin x-(x-1)^2, \qquad x\in[0,2].$$

We count zeros of $f$.

Check some values:

$$f(0)=\sin 0-1=-1<0,$$ $$f(1)=\sin 1-0=\sin 1>0,$$ $$f(2)=\sin 2-1<1-1=0.$$

So by continuity:

• one root lies in $(0,1)$,
• one root lies in $(1,2)$.

Now we show there are no more than these two.

Differentiate:

$$f'(x)=\cos x-2(x-1)=\cos x-2x+2.$$

And

$$f''(x)=-\sin x-2<0 \quad \text{for } x\in[0,2].$$

So $f$ is strictly concave on $[0,2]$. Hence $f'$ is strictly decreasing, and therefore $f$ can cross the $x$-axis at most twice.

Since we already found at least two roots, it follows that there are exactly two roots.

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5. Number of solutions

Therefore, the number of solutions in $[-\pi,\pi]$ is

$$\boxed{2}.$$

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6. Comparison with stored answer

Stored correct answer: $2$

Our derived answer is also $2$, so they agree.