Animated Solution for Mathematics - Trigonometry: The number of values of $x$ in the interval $[0,3\pi ]$ satisfying the equation $2{\sin }^{2}x+5\sin x-3=0$ is

Welcome, future engineer! Today, we are going to peel back the layers of a trigonometric problem that, at first glance, might look like a standard identity exercise.

The equation is a classic example of a 'wolf in sheep's clothing.' It is a quadratic equation masquerading as trigonometry.

When you see a squared term and a linear term of the same function, your first instinct should always be to simplify. Let us define a new variable, $t=\sin x$.

Suddenly, the complexity vanishes, and we are left with the familiar:

Now, we enter the realm of algebra. We need to split the middle term to factorize this quadratic.

We are looking for two numbers that multiply to $2\times (-3)=-6$ and add to $5$. Those numbers are $6$ and $-1$.

So, we rewrite the equation as:

By grouping terms, we get $2t(t+3)-1(t+3)=0$, which leads us to the beautiful factorization:

This gives us two potential roots: $t=\frac{1}{2}$ and $t=-3$.

Here is where many students stumble. We must ask ourselves: does $t=-3$ make sense?

Since $t=\sin x$, and we know the range of the sine function is strictly $[-1,1]$, the value $-3$ is impossible. It is a 'ghost root'—a mathematical artifact that has no place in the physical world of angles.

We discard it immediately. We are left with the elegant, singular requirement:

Now, let us step into the shoes of an observer watching the sine wave oscillate. We are restricted to the interval $[0,3\pi ]$.

Imagine the graph of $y=\sin x$ starting at $0$, rising to $1$ at $\frac{\pi }{2}$, falling back to $0$ at $\pi$, dipping to $-1$ at $\frac{3\pi }{2}$, and returning to $0$ at $2\pi$. That is one full cycle.

But we are not done! We must continue for another half-cycle, from $2\pi$ to $3\pi$.

We draw a horizontal line at $y=\frac{1}{2}$. Where does it cut the wave?

In the first cycle $[0,2\pi ]$, the line intersects the curve twice: once in the first quadrant at $x=\frac{\pi }{6}$ and once in the second quadrant at $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.

But wait, the interval goes up to $3\pi$. Because the sine function is periodic with a period of $2\pi$, we can simply add $2\pi$ to our previous solutions to find the intersections in the next cycle.

Thus, $x=2\pi +\frac{\pi }{6}=\frac{13\pi }{6}$ and $x=2\pi +\frac{5\pi }{6}=\frac{17\pi }{6}$.

We check our boundaries: is $\frac{17\pi }{6}$ less than $3\pi$? Since $3\pi =\frac{18\pi }{6}$, yes, it is!

We have successfully captured all four solutions:

Counting them up, we find exactly 4 values of $x$ that satisfy the equation.

The beauty of this problem lies not just in the algebra, but in the synthesis of algebraic constraints and geometric visualization. You didn't just solve an equation; you mapped the behavior of a wave across a specific domain.

Keep this mindset—always look for the constraint, always visualize the graph, and never let a 'ghost root' distract you from the truth. You are doing great!