Animated Solution for Mathematics - Trigonometry: Number of solutions of the equation $\tan x+\sec x=2\cos x$ lying in the interval $[0,2\pi ]$ is :

Welcome, future engineer. Today, we are not just solving an equation; we are embarking on a journey of logical precision.

The equation $\tan x+\sec x=2\cos x$ might look simple, but it hides a classic trap that has caught many students off guard. Let us break it down, step by step, and uncover the beauty hidden within.

The Setup and the Hidden Constraint

We are given the equation $\tan x+\sec x=2\cos x$ within the interval $[0,2\pi ]$. Before we even touch our pens to paper, we must pause.

In the world of trigonometry, functions like $\tan x$ and $\sec x$ are not defined everywhere. Specifically, they involve a division by $\cos x$.

This means that whenever $\cos x=0$, our equation effectively ceases to exist. This is our first, silent constraint: $\cos x eq 0$. Keep this in your mental toolkit; it will be the judge and jury for our final answers.

The Transformation

When you see a mix of tangent, secant, and cosine, the most powerful strategy is unification. We want to speak a single language.

Let us rewrite the left-hand side using the fundamental definitions:

Substituting these into our equation, the left-hand side becomes $\frac{\sin x+1}{\cos x}$. Now, our equation looks like this:

The Quadratic Bridge

To eliminate the fraction, we multiply both sides by $\cos x$. This gives us $1+\sin x=2{\cos }^{2}x$.

We are almost there, but we still have a mix of sine and cosine. To solve this, we need a single trigonometric ratio.

We invoke the most famous identity in trigonometry: ${\cos }^{2}x=1-{\sin }^{2}x$. Substituting this in, we get:

Expanding the right side, we get $1+\sin x=2-2{\sin }^{2}x$. Bringing all terms to one side, we arrive at a beautiful, standard quadratic equation in terms of $\sin x$:

The Factorization and the Trap

Now, we factorize. We split the middle term: $2{\sin }^{2}x+2\sin x-\sin x-1=0$.

This simplifies to $(2\sin x-1)(\sin x+1)=0$. This gives us two possible cases for $\sin x$:

1. $2\sin x-1=0\Rightarrow \sin x=\frac{1}{2}$

2. $\sin x+1=0\Rightarrow \sin x=-1$

For $\sin x=\frac{1}{2}$, the solutions in $[0,2\pi ]$ are $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$. For $\sin x=-1$, the solution is $x=\frac{3\pi }{2}$.

The Final Verification

Here is where the master educator reminds you: never trust a solution until you verify it against the domain. We established earlier that $\cos x eq 0$.

Let us check our candidates:

- At $x=\frac{\pi }{6}$, $\cos x = \frac{\sqrt{3}}{2} eq 0$. (Valid)

- At $x=\frac{5\pi }{6}$, $\cos x = -\frac{\sqrt{3}}{2} eq 0$. (Valid)

- At $x=\frac{3\pi }{2}$, $\cos x=0$. (Invalid!)

Because $\cos x=0$ at $x=\frac{3\pi }{2}$, the original equation $\tan x+\sec x=2\cos x$ is undefined at this point. Thus, $x=\frac{3\pi }{2}$ is an extraneous solution and must be rejected.

We are left with exactly two valid solutions: $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$.

The total number of solutions is 2.

This problem teaches us that in JEE Advanced, the math is only half the battle; the other half is maintaining the integrity of the domain. Keep this vigilance, and you will conquer any problem they throw at you!