Animated Solution for Mathematics - Trigonometry: The number of solutions of equation $(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x=-\frac{4}{1+\sqrt{3}},x\in [-2\pi ,\frac{5\pi }{2}]$ is

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of irrational numbers and trigonometric ratios.

When you see an equation like $(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x=-\frac{4}{1+\sqrt{3}}$, do not panic. Instead, take a deep breath and look for the structure. We are going to peel back the layers of this problem together.

The Clean-Up

Our first instinct should always be to simplify. Look at the right-hand side: $-\frac{4}{1+\sqrt{3}}$. That irrational denominator is designed to make you hesitate.

Let us rationalize it by multiplying the numerator and denominator by the conjugate, $\sqrt{3}-1$:

Suddenly, the equation looks much friendlier. We have transformed a daunting fraction into a clean constant, which is the first step in gaining control over the problem.

The Transformation

Now, look at the left-hand side: $(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x$. We have a mix of $\sin x$ and ${\cos }^{2}x$. We need a common language, so we use the identity ${\cos }^{2}x=1-{\sin }^{2}x$:

When we expand this, the term $-2\sqrt{3}$ appears on both sides of the equation and cancels out perfectly. We are left with a clean quadratic equation in terms of $\sin x$:

The Filter

Let us treat $\sin x$ as a variable, $s$. We factorize $2\sqrt{3}{s}^2+4s-\sqrt{3}s-2=0$ by grouping:

This gives us two potential roots: $\sin x=\frac{1}{2}$ or $\sin x=-\frac{2}{\sqrt{3}}$.

Here is where the trap lies. The value $-\frac{2}{\sqrt{3}}$ is approximately $-1.15$. Since the sine function is bounded between $-1$ and $1$, this root is physically impossible and must be rejected.

The Visual Journey

We are tasked with finding the number of solutions in the interval $x\in [-2\pi ,\frac{5\pi }{2}]$. We rely on the geometry of the sine wave:

1. The Negative Cycle ($[-2\pi ,0]$): The sine wave completes one full negative cycle. The line $y=\frac{1}{2}$ cuts the wave twice, at $x=-2\pi +\frac{\pi }{6}=-\frac{11\pi }{6}$ and $x=-2\pi +\frac{5\pi }{6}=-\frac{7\pi }{6}$.

2. The Positive Cycle ($[0,2\pi ]$): In the standard positive range, the sine wave hits $\frac{1}{2}$ at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$. This provides another 2 solutions.

3. The Final Stretch ($[2\pi ,\frac{5\pi }{2}]$): In the interval from $2\pi$ to $2.5\pi$, the sine wave rises from $0$ to $1$. It crosses the line $y=\frac{1}{2}$ exactly once at $x=2\pi +\frac{\pi }{6}=\frac{13\pi }{6}$.

Adding the solutions from each interval, we get $2+2+1=5$.

We have navigated the algebra, avoided the domain traps, and visualized the wave to find exactly 5 solutions. This problem was not just about solving for $x$; it was about maintaining clarity amidst complexity.