Animated Solution for Mathematics - Trigonometry: If $\theta \in [-\frac{7\pi }{6},\frac{4\pi }{3}]$, then the number of solutions of $\sqrt{3}{\mathrm{cosec}}^2\theta -2(\sqrt{3}-1)\mathrm{cosec}\theta -4=0$, is equal to

Welcome, JEE warriors. Today, we are not just solving an equation; we are peeling back the layers of a trigonometric mystery.

Trigonometry often hides its true nature behind complex functions like $\mathrm{cosec}\theta$. But look closely at the structure:

Does it not look like a quadratic equation $a{x}^2+bx+c=0$ wearing a disguise? This is the first lesson of the JEE: pattern recognition.

By letting $x=\mathrm{cosec}\theta$, we strip away the trigonometric complexity and reveal the algebraic core:

Suddenly, the problem feels manageable, doesn't it?

Now, we enter the realm of the discriminant. Many students fear the quadratic formula, but it is your most loyal ally.

We identify our coefficients: $a=\sqrt{3}$, $b=-2(\sqrt{3}-1)$, and $c=-4$. The discriminant $D=b^2-4ac$ is the heartbeat of this equation.

When we calculate $D=[-2(\sqrt{3}-1){]}^2-4(\sqrt{3})(-4)$, we are looking for a perfect square. Expanding this, we get:

Factoring out a $4$, we find $4(4+2\sqrt{3})$. And here is the magic: $4+2\sqrt{3}$ is exactly $(\sqrt{3}+1{)}^2$.

The discriminant is $4(\sqrt{3}+1{)}^2$. When you see a perfect square in the discriminant, you know you are on the right path. It is the universe rewarding your patience.

With the roots $x=2$ and $x=-\frac{2}{\sqrt{3}}$ in hand, we return to the world of trigonometry.

We have $\mathrm{cosec}\theta =2$, which implies $\sin \theta =\frac{1}{2}$, and $\mathrm{cosec}\theta =-\frac{2}{\sqrt{3}}$, which implies $\sin \theta =-\frac{\sqrt{3}}{2}$.

Now, we must map these to the interval $[-\frac{7\pi }{6},\frac{4\pi }{3}]$. Imagine the sine wave undulating across your page.

We draw the horizontal lines $y=\frac{1}{2}$ and $y=-\frac{\sqrt{3}}{2}$. The intersections are not just numbers; they are points on the wave.

For $\sin \theta =\frac{1}{2}$, we find solutions at $\theta =-\frac{7\pi }{6}$, $\frac{\pi }{6}$, and $\frac{5\pi }{6}$. Notice how the boundary $-\frac{7\pi }{6}$ is included? That is a critical catch.

For $\sin \theta =-\frac{\sqrt{3}}{2}$, we find solutions at $\theta =-\frac{2\pi }{3}$, $-\frac{\pi }{3}$, and $\frac{4\pi }{3}$. Again, the boundary $\frac{4\pi }{3}$ is included.

Counting them up, we have $3+3=6$ solutions. You have successfully navigated the trap.

Remember, in the JEE, the difference between a good rank and a great rank is often found in these boundary checks. The total number of solutions is 6.