Animated Solution for Mathematics - Trigonometry: The number of distinct solutions of the equation $\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2$ in the interval $[0,2\pi ]$ is ____.

Imagine you are standing before a massive, intimidating structure. It looks complex, perhaps even impossible to dismantle.

That is exactly how this equation feels:

It is a dense thicket of powers and trigonometric functions. But in the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to peel back these layers and reveal the simple truth hidden underneath.

Taming the Power of Four

Let us start with the most manageable parts: ${\sin }^{4}x+{\cos }^{4}x$. We do not want to deal with fourth powers if we can avoid them.

We recall the algebraic identity $a^4+b^4=(a^2+b^2{)}^2-2a^2{b}^2$. By setting $a=\sin x$ and $b=\cos x$, we transform this into $({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x$.

Since ${\sin }^{2}x+{\cos }^{2}x=1$, this simplifies to $1-2{\sin }^{2}x{\cos }^{2}x$. To make this even cleaner, we use the double angle identity $\sin 2x=2\sin x\cos x$, which implies ${\sin }^{2}2x=4{\sin }^{2}x{\cos }^{2}x$.

Thus, our expression becomes $1-\frac{1}{2}{\sin }^{2}2x$. We have successfully tamed the first beast!

The Power of Six

Now, let us tackle ${\sin }^{6}x+{\cos }^{6}x$. This looks scarier, but we can treat it as a sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$, where $a={\sin }^{2}x$ and $b={\cos }^{2}x$.

The first bracket, $({\sin }^{2}x+{\cos }^{2}x)$, is just $1$. The second bracket is $({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x)$.

We already know ${\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x$. Substituting this in, we get $1-2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x$, which simplifies to $1-3{\sin }^{2}x{\cos }^{2}x$.

Again, using our double angle trick, this becomes $1-\frac{3}{4}{\sin }^{2}2x$.

The Great Cancellation

With our simplified components, let us return to the original equation:

Look closely at the constants. We have $1+1=2$ on the left, and a $2$ on the right. They cancel out perfectly!

We are left with:

Combining the sine terms, $-\frac{1}{2}-\frac{3}{4}=-\frac{5}{4}$. The equation is now:

The Final Reveal

Dividing by $\frac{5}{4}$, we get ${\cos }^{2}2x-{\sin }^{2}2x=0$. This is the classic identity $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$.

With $\theta =2x$, our equation is simply $\cos 4x=0$. The monster has vanished, replaced by a simple, elegant trigonometric equation.

We need to find the number of solutions for $x\in [0,2\pi ]$, which means $4x\in [0,8\pi ]$. The cosine function is zero at odd multiples of $\frac{\pi }{2}$.

In the range $[0,8\pi ]$, there are exactly 8 such points:

Each corresponds to a unique value of $x$. Thus, there are exactly 8 distinct solutions. You have conquered the problem!