Animated Solution for Mathematics - Trigonometry: Number of solutions of $\sqrt{3}\cos 2\theta +8\cos \theta +3\sqrt{3}=0,\theta \in [-3\pi ,2\pi ]$ is:

Unifying the Angles

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that might look like a chaotic mess of square roots and double angles, but beneath the surface, it is a beautiful, structured puzzle.

Our mission is to find the number of solutions for the equation $\sqrt{3}\cos 2\theta +8\cos \theta +3\sqrt{3}=0$ within the interval $\theta \in [-3\pi ,2\pi ]$.

The Language of Unification

When you see an equation with mixed angles like $2\theta$ and $\theta$, your first instinct should be to unify them. We cannot easily solve an equation that speaks two different languages.

We need a common ground. This is where the double-angle identity for cosine comes to our rescue: $\cos 2\theta =2{\cos }^2\theta -1$.

By substituting this into our original equation, we transform the entire expression into a single language: the language of $\cos \theta$. The equation becomes:

The Quadratic Transformation

Now, let us expand and simplify. Distributing the $\sqrt{3}$ gives us $2\sqrt{3}{\cos }^2\theta -\sqrt{3}+8\cos \theta +3\sqrt{3}=0$.

Combining the constant terms, $-\sqrt{3}+3\sqrt{3}$, yields $2\sqrt{3}$. Our equation is now:

To make our lives easier, we divide by $2$, resulting in $\sqrt{3}{\cos }^2\theta +4\cos \theta +\sqrt{3}=0$. If we let $x=\cos \theta$, we see the elegant structure of a quadratic equation:

The Reality Check

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find the roots:

This gives us two potential values: $x=-\frac{1}{\sqrt{3}}$ and $x=-\sqrt{3}$.

Here is the crucial moment of truth. We must remember that $x=\cos \theta$, and the cosine function is bounded by $[-1,1]$.

Since $-\sqrt{3}\approx -1.732$, it is impossible for $\cos \theta$ to equal this value. We reject it immediately. We are left with the single, valid condition: $\cos \theta =-\frac{1}{\sqrt{3}}$.

The Graphical Odyssey

Now, we must count the solutions in the interval $[-3\pi ,2\pi ]$. Instead of solving for $\theta$, we visualize the intersection of the curve $y=\cos \theta$ and the horizontal line $y=-\frac{1}{\sqrt{3}}$.

In the interval $[0,2\pi ]$, the cosine curve completes one full cycle. Since the line $y=-\frac{1}{\sqrt{3}}$ is negative, it intersects the curve twice (in the second and third quadrants). That is 2 solutions.

In the interval $[-2\pi ,0]$, the cosine curve completes another full cycle. The symmetry of the cosine function ensures it intersects the line twice here as well. That is another 2 solutions.

Finally, in the interval $[-3\pi ,-2\pi ]$, we have half a cycle. The curve dips from $-1$ to $1$. The line $y=-\frac{1}{\sqrt{3}}$ cuts through this dip exactly once. That is 1 solution.

Adding them all up, $2+2+1=5$. We have arrived at our destination! The total number of solutions is 5.