Animated Solution for Mathematics - Trigonometry: Let $S=\{\theta \in [-2\pi ,2\pi ]:2{\cos }^2\theta +3\sin \theta =0\}$. Then the sum of the elements of $S$ is

Welcome, future engineer! Today, we are going to unravel a problem that might look like a simple algebraic exercise, but it is actually a beautiful dance of trigonometric identities and geometric intuition.

We are tasked with solving the equation $2{\cos }^2\theta +3\sin \theta =0$ within the interval $\theta \in [-2\pi ,2\pi ]$.

At first glance, you might feel a bit of friction. We have a cosine term and a sine term living in the same house, and they don't quite speak the same language. To solve this, we need a translator.

That translator is the fundamental identity: ${\sin }^2\theta +{\cos }^2\theta =1$. By rearranging this, we can express ${\cos }^2\theta$ as $1-{\sin }^2\theta$.

By substituting this into our original equation, we transform the problem into:

Suddenly, the complexity vanishes, and we are left with a single, unified language: the language of sine.

Now that we have our equation in terms of $\sin \theta$, let's expand it: $2-2{\sin }^2\theta +3\sin \theta =0$.

If we multiply the entire equation by $-1$ to make the leading coefficient positive, we get:

Does this look familiar? It should! It is a classic quadratic equation. If you find it intimidating, just let $x=\sin \theta$. Now you are solving $2x^2-3x-2=0$.

We can split the middle term: $2x^2-4x+x-2=0$. Grouping these gives us $2x(x-2)+1(x-2)=0$, which factors beautifully into $(2x+1)(x-2)=0$.

Substituting $\sin \theta$ back in, we get $(2\sin \theta +1)(\sin \theta -2)=0$. This is the moment of truth. We have two potential paths: $\sin \theta =-\frac{1}{2}$ or $\sin \theta =2$.

The Range Trap

Here is where many students stumble. We have two roots, but are they both physically possible?

Think back to the unit circle. The sine function is the projection of a point on the unit circle onto the y-axis. Because the circle is bounded by a radius of 1, the value of $\sin \theta$ is strictly confined to the interval $[-1,1]$.

Therefore, the root $\sin \theta =2$ is a mathematical phantom—it simply cannot exist. We must reject it firmly. We are left with the single, valid condition: $\sin \theta =-\frac{1}{2}$.

Now, we must find all values of $\theta$ in the interval $[-2\pi ,2\pi ]$ such that $\sin \theta =-\frac{1}{2}$. Let's visualize the unit circle.

We know that $\sin (\frac{\pi }{6})=\frac{1}{2}$. Since we need $\sin \theta =-\frac{1}{2}$, we are looking for angles in the third and fourth quadrants where the sine is negative.

In the positive interval $[0,2\pi ]$, the third quadrant angle is $\pi +\frac{\pi }{6}=\frac{7\pi }{6}$, and the fourth quadrant angle is $2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$.

Now, what about the negative interval $[-2\pi ,0]$? We simply move clockwise. The fourth quadrant angle is $-\frac{\pi }{6}$, and the third quadrant angle is $-\pi +\frac{\pi }{6}=-\frac{5\pi }{6}$.

We have found our four solutions: $\theta \in \{-\frac{5\pi }{6},-\frac{\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\}$.

We have reached the final stage of our journey. We need the sum of these elements. Let's add them up:

Since the denominators are identical, we simply sum the numerators: $-5\pi -\pi +7\pi +11\pi =12\pi$.

Dividing by the common denominator of 6, we get:

The elegance of this result is a testament to the symmetry of the trigonometric functions. You have successfully navigated the identity, the quadratic, the range trap, and the unit circle. The final answer is $2\pi$.