The number of solutions of equation `tanx+secx=2cosx` lying in the interval `[0, 2pi]` is

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To find the number of solutions of the equation \( \tan x + \sec x = 2 \cos x \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the equation using sine and cosine
The given equation is:
\[ \tan x + \sec x = 2 \cos x \]
We can express \( \tan x \) and \( \sec x \) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x} \]
Substituting these into the equation gives:
\[ \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x \]

### Step 2: Combine the left-hand side
Combining the terms on the left-hand side:
\[ \frac{\sin x + 1}{\cos x} = 2 \cos x \]
Multiplying both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \sin x + 1 = 2 \cos^2 x \]

### Step 3: Use the Pythagorean identity
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the equation:
\[ \sin x + 1 = 2(1 - \sin^2 x) \]
Expanding the right-hand side:
\[ \sin x + 1 = 2 - 2 \sin^2 x \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 2 \sin^2 x + \sin x + 1 - 2 = 0 \]
This simplifies to:
\[ 2 \sin^2 x + \sin x - 1 = 0 \]

### Step 5: Factor the quadratic equation
Now we will factor the quadratic equation:
\[ 2 \sin^2 x + 2 \sin x - \sin x - 1 = 0 \]
Factoring gives:
\[ (2 \sin x - 1)(\sin x + 1) = 0 \]

### Step 6: Solve for \( \sin x \)
Setting each factor to zero gives us:
1. \( 2 \sin x - 1 = 0 \) → \( \sin x = \frac{1}{2} \)
2. \( \sin x + 1 = 0 \) → \( \sin x = -1 \)

### Step 7: Find solutions in the interval \( [0, 2\pi] \)
1. For \( \sin x = \frac{1}{2} \):
- The solutions in the interval \( [0, 2\pi] \) are:
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)

2. For \( \sin x = -1 \):
- The solution in the interval \( [0, 2\pi] \) is:
- \( x = \frac{3\pi}{2} \)

### Step 8: Check if \( x = \frac{3\pi}{2} \) is a solution
Substituting \( x = \frac{3\pi}{2} \) into the original equation:
\[ \tan\left(\frac{3\pi}{2}\right) + \sec\left(\frac{3\pi}{2}\right) = 2 \cos\left(\frac{3\pi}{2}\right) \]
Here, \( \tan\left(\frac{3\pi}{2}\right) \) is undefined, so \( x = \frac{3\pi}{2} \) is not a valid solution.

### Conclusion
Thus, the valid solutions in the interval \( [0, 2\pi] \) are:
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)

Therefore, the number of solutions is **2**.