Animated Solution for Mathematics - Trigonometry: Let the set of all $a\in \mathbb{R}$ such that the equation $\cos 2x+a\sin x=2a-7$ has a solution be $[p,q]$ and $r=\tan 9^{\circ }-\tan 27^{\circ }-\frac{1}{\cot 63^{\circ }}+\tan 81^{\circ }$, then $pqr$ is equal to

A JEE Advanced Journey

Welcome, future engineers. Today, we are going to dissect a problem that at first glance might seem like a chaotic mess of trigonometric functions, but beneath the surface, it is a beautifully structured puzzle.

In the world of JEE Advanced, we don't just solve equations; we look for the underlying harmony. Let us embark on this journey together.

The Unification

We begin with the equation $\cos 2x+a\sin x=2a-7$. The first thing that should strike you is the duality: we have a $\cos 2x$ term and a $\sin x$ term.

To make progress, we must translate them into a common tongue. The double angle identity is our bridge:

By substituting this into our equation, we get:

Now, everything is in terms of $\sin x$. This is the first step toward clarity.

The Art of Factoring

With our equation transformed into $2{\sin }^{2}x-a\sin x+2a-8=0$, we are looking at a quadratic in $\sin x$. Many students would immediately reach for the quadratic formula, but in JEE, there is often a more elegant path.

Let us group the terms:

Notice the difference of squares in the first term, where ${\sin }^{2}x-4=(\sin x-2)(\sin x+2)$. This reveals a common factor of $(\sin x-2)$.

Factoring it out, we get:

This is the 'Aha!' moment. Since $\sin x$ is trapped between $-1$ and $1$, it can never be $2$. Thus, the factor $(\sin x-2)$ is never zero.

We are left with $2(\sin x+2)-a=0$, or simply:

The Range of Possibilities

Now that we have $a=2\sin x+4$, finding the range $[p,q]$ is straightforward. Since $-1\le \sin x\le 1$, we multiply by $2$ to get $-2\le 2\sin x\le 2$.

Adding $4$ to the inequality, we obtain:

Thus, $a\in [2,6]$, giving us $p=2$ and $q=6$. The first part of our puzzle is complete.

The Elegance of Identities

Next, we tackle the expression $r=\tan 9^{\circ }-\tan 27^{\circ }-\frac{1}{\cot 63^{\circ }}+\tan 81^{\circ }$. Using complementary angles, we know $\tan 81^{\circ }=\cot 9^{\circ }$ and $\frac{1}{\cot 63^{\circ }}=\tan 63^{\circ }=\cot 27^{\circ }$.

The expression becomes:

Using the identity $\tan \theta +\cot \theta =\frac{2}{\sin 2\theta }$, we get:

Substituting the standard values $\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ }=\frac{\sqrt{5}+1}{4}$, the expression simplifies beautifully to $r=4$.

Finally, the product is:

We have arrived at the destination. Remember, in mathematics, as in life, complexity is often just a mask for a deeper, simpler truth.