Animated Solution for Mathematics - Trigonometry: The number of solutions of ${\sin }^{2}x+(2+2x-x^2)\sin x-3(x-1{)}^2=0$, where $-\pi \le x\le \pi$, is________

The given equation is:

This equation presents a collision between trigonometric and algebraic functions. To solve it, we must look for hidden structures within the coefficients.

Focus on the middle coefficient: $2+2x-x^2$. By rearranging the terms, we can express this as:

Substituting this back into the original equation, we obtain:

This transformation reveals that $(x-1{)}^2$ is the central component of the equation.

To simplify the expression, let $a=\sin x$ and $b=(x-1{)}^2$. The equation now takes the form of a quadratic:

Expanding and grouping the terms, we get:

This yields two potential paths: $a=b$ or $a=-3$.

Consider the path $a=-3$. Substituting back, we have $\sin x=-3$.

Since the range of the sine function is restricted to $[-1,1]$, the equation $\sin x=-3$ has no real solutions. We discard this path as a mathematical mirage.

We are left with the condition $\sin x=(x-1{)}^2$. We must find the number of intersections between $y=\sin x$ and $y=(x-1{)}^2$ within the interval $[-\pi ,\pi ]$.

In the negative domain $[-\pi ,0]$, $\sin x$ is non-positive, while $(x-1{)}^2$ is strictly positive. Thus, there are no intersections here.

In the positive domain $[0,\pi ]$:

1. At $x=0$, $\sin (0)=0$ and $(0-1{)}^2=1$. Here, the parabola is above the sine wave. 2. At $x=1$, $\sin (1)\approx 0.84$ and $(1-1{)}^2=0$. The sine wave is now above the parabola. By the Intermediate Value Theorem, there is at least one solution in $(0,1)$. 3. At $x=\pi$, $\sin (\pi )=0$ and $(\pi -1{)}^2\approx 4.58$. The parabola is once again above the sine wave. By the Intermediate Value Theorem, there is at least one solution in $(1,\pi )$.

The Victory

By systematically analyzing the domain, we have determined that there are no solutions in the negative region and exactly two solutions in the positive region.

The total number of solutions is 2.