Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x}$ in the interval $[0,2\pi ]$ is

Defining the Domain

In any equation involving fractions, the denominator is the boss. Here, we have the term $\frac{1}{\sin x}$.

This immediately tells us that $\sin x$ cannot be zero. If $\sin x$ were zero, the entire equation would explode into undefined territory.

Looking at our interval $[0,2\pi ]$, we know that $\sin x=0$ at $x=0,\pi ,$ and $2\pi$. Our valid playground is strictly the open intervals $(0,\pi )$ and $(\pi ,2\pi )$.

Case Analysis

The modulus function $|\cot x|$ is a shape-shifter. It behaves differently depending on whether the input is positive or negative, forcing us to split our journey into two distinct paths.

Assume $\cot x\ge 0$. In this scenario, the modulus simply vanishes, and the equation becomes:

If we subtract $\cot x$ from both sides, we are left with $0=\frac{1}{\sin x}$. Since a fraction with a constant numerator of $1$ can never equal zero, this path leads to a dead end.

Assume $\cot x<0$. Now, the modulus opens with a negative sign, transforming our equation into:

Moving the $\cot x$ to the left side, we obtain:

Now, let us translate this into the universal language of sine and cosine. We know that $\cot x=\frac{\cos x}{\sin x}$.

Substituting this, we get:

Because we have already established that $\sin x eq 0$, we can confidently multiply both sides by $\sin x$. This leaves us with the simplified equation:

We need to find where $\cos x=-\frac{1}{2}$ in the interval $[0,2\pi ]$. On the unit circle, this occurs at two points: $x=\frac{2\pi }{3}$ and $x=\frac{4\pi }{3}$.

We must return to our original assumption for Path 2, which required $\cot x<0$.

At $x=\frac{2\pi }{3}$, we are in the second quadrant where cotangent is negative. This solution is valid.

At $x=\frac{4\pi }{3}$, we are in the third quadrant where cotangent is positive. This violates our condition and must be rejected.

Thus, after navigating the traps and verifying our constraints, we find that there is exactly one valid solution: