Animated Solution for Mathematics - Trigonometry: The number of elements in the set $S=\{\theta \in [0,2\pi ]:3{\cos }^4\theta -5{\cos }^2\theta -2{\sin }^6\theta +2=0\}$ is

Welcome, future engineer. Today, we stand before a problem that might initially seem like a chaotic mess of powers and ratios:

When you see an equation like this in a JEE Advanced paper, your first instinct might be panic. But take a deep breath. In the world of competitive mathematics, complexity is often just a mask for symmetry. Our goal is to strip away that mask.

The Language of Sine

We cannot solve an equation that speaks two different languages—sine and cosine. We must unify them. The most elegant path here is to translate everything into the language of sine.

We know the fundamental identity ${\cos }^2\theta =1-{\sin }^2\theta$. By substituting this into our equation, we replace the cosine terms. The equation transforms into:

Suddenly, the trigonometric chaos begins to settle. We are now dealing with a single variable, $\sin \theta$.

The Algebraic Transformation

Trigonometry is often just algebra in disguise. To make this clear, let us introduce a dummy variable. Let $t={\sin }^2\theta$.

Because $\sin \theta$ is bounded between $-1$ and $1$, our new variable $t$ must reside strictly within the interval $[0,1]$. This constraint is our safety net; it will catch any extraneous roots that might try to trick us later.

Substituting $t$ into our equation, we get:

Now, we expand. The term $(1-t{)}^2$ becomes $1-2t+t^2$. Distributing the constants, we carefully simplify the expression to:

Combining like terms, we arrive at the beautiful, clean cubic equation:

The Cubic Challenge

We have arrived at the heart of the problem. Factoring $2t^3-3t^2+t=0$ is straightforward. We pull out the common factor $t$, leaving us with:

The quadratic part, $2t^2-3t+1$, factors further into $(2t-1)(t-1)=0$. This gives us three distinct values for $t$:

All three values fall within our valid range of $[0,1]$, so we proceed with confidence.

Mapping Back to the Unit Circle

Now, we return to our original variable, $\theta$. We have three cases to solve for $\theta \in [0,2\pi ]$.

Case 1: ${\sin }^2\theta =0$. This implies $\sin \theta =0$. On the unit circle, the sine function is zero at $0$, $\pi$, and $2\pi$. That is 3 solutions.

Case 2: ${\sin }^2\theta =1$. This implies $\sin \theta =\pm 1$. This occurs at the top and bottom of the unit circle: $\pi /2$ and $3\pi /2$. That is 2 solutions.

Case 3: ${\sin }^2\theta =1/2$. This implies $\sin \theta =\pm 1/\sqrt{2}$. These are the classic $45^{\circ }$ reference angles in all four quadrants: $\pi /4$, $3\pi /4$, $5\pi /4$, and $7\pi /4$. That is 4 solutions.

Summing them up, $3+2+4=9$. We have successfully navigated the complexity and found exactly 9 solutions.