Animated Solution for Mathematics - Trigonometry: The number of solution of $\tan x+\sec x=2\cos x$ in $[0,2\pi )$ is

Welcome, future engineers! Today, we are going to dissect a problem that is a classic in the JEE Advanced repertoire. It is a problem that tests not just your algebraic manipulation, but your mathematical maturity.

We are looking for the number of solutions to the equation $\tan x+\sec x=2\cos x$ within the interval $[0,2\pi )$. At first glance, this looks like a straightforward trigonometric equation. But as you will soon see, the beauty of this problem lies in the hidden traps.

Before we even touch the algebra, we must respect the domain. In mathematics, functions are like contracts; they only hold true where they are defined.

Here, we have $\tan x$ and $\sec x$. Recall that $\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$. Both of these functions share a common enemy: $\cos x=0$.

If $\cos x$ is zero, the denominator vanishes, and the function explodes into infinity. On our unit circle, $\cos x=0$ at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$. These are our 'danger zones.'

We must mark these points immediately. Any solution we find that lands on these points is an impostor—an extraneous solution—and must be rejected.

Now that we have established our boundaries, let's simplify. The most reliable strategy in trigonometry is to convert everything into the fundamental language of $\sin x$ and $\cos x$.

Our equation becomes:

Since the left side shares a common denominator, we can combine them:

Now, we can multiply both sides by $\cos x$ to clear the fraction. We can safely do this because we have already explicitly excluded the values where $\cos x=0$. This gives us:

We are now looking at $\sin x+1=2{\cos }^{2}x$. We have a mix of sine and cosine. To solve this, we need a single trigonometric ratio.

We use the Pythagorean identity: ${\cos }^{2}x=1-{\sin }^{2}x$. Substituting this into our equation, we get:

Expanding the right side, we get $\sin x+1=2-2{\sin }^{2}x$. Let's bring everything to one side to form a standard quadratic equation:

This is the heart of the problem. We are now solving a quadratic in terms of $\sin x$. Let $y=\sin x$. The equation is $2y^2+y-1=0$.

Factoring this, we get $(2y-1)(y+1)=0$. This gives us two cases: 1. $2\sin x-1=0\Rightarrow \sin x=\frac{1}{2}$ 2. $\sin x+1=0\Rightarrow \sin x=-1$

Now, we find the angles. 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 $x=\frac{3\pi }{2}$.

Now, we return to our domain constraints from Phase 1. We said $\cos x eq 0$. Let's check our candidates.

At $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$, $\cos x$ is non-zero. These are valid. But at $x=\frac{3\pi }{2}$, $\cos (\frac{3\pi }{2})=0$. This is our danger zone!

We must reject $x=\frac{3\pi }{2}$. Thus, 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. Remember, in JEE, the math is only half the battle; the other half is vigilance. Keep your eyes on the domain, and you will never be trapped.