solve equation `2 tan^(-1)(cosx)=tan^(-1)(2cosecx)`

Explanation

To solve the equation \( 2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x) \), we can follow these steps:

### Step 1: Use the double angle formula for tangent
We know that \( 2 \tan^{-1}(y) = \tan^{-1}\left(\frac{2y}{1 - y^2}\right) \). Therefore, we can rewrite the left side of the equation:
\[ 2 \tan^{-1}(\cos x) = \tan^{-1\left(\frac{2 \cos x}{1 - \cos^2 x}\right) \]

### Step 2: Simplify the expression
Recall that \( 1 - \cos^2 x = \sin^2 x \). Thus, we can simplify:
\[ \tan^{-1}\left(\frac{2 \cos x}{\sin^2 x}\right) = \tan^{-1}(2 \csc x) \]

### Step 3: Set the arguments equal
Since \( \tan^{-1}(A) = \tan^{-1}(B) \) implies \( A = B \) (as long as both are in the principal range of the arctangent function), we can set the arguments equal:
\[ \frac{2 \cos x}{\sin^2 x} = 2 \csc x \]

### Step 4: Rewrite \( \csc x \)
Recall that \( \csc x = \frac{1}{\sin x} \), so we can rewrite the right side:
\[ \frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 2 \cos x \cdot \sin x = 2 \sin^2 x \]

### Step 6: Simplify the equation
Dividing both sides by 2 (assuming \( \sin x \neq 0 \)):
\[ \cos x \cdot \sin x = \sin^2 x \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \cos x \cdot \sin x - \sin^2 x = 0 \]
Factoring out \( \sin x \):
\[ \sin x (\cos x - \sin x) = 0 \]

### Step 8: Solve for \( x \)
This gives us two cases:
1. \( \sin x = 0 \)
2. \( \cos x - \sin x = 0 \)

From \( \sin x = 0 \), we have:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]

From \( \cos x - \sin x = 0 \), we have:
\[ \cos x = \sin x \implies x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Final Solution
Thus, the solutions to the equation are:
\[ x = n\pi \quad \text{and} \quad x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]