Find the number of solution of ` tan x + sec x =2 cos x ` in `[0,2pi]`

Answer

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

### Step 1: Rewrite the equation
The given equation is:
\[ \tan x + \sec x = 2 \cos x \]
We can rewrite \(\tan x\) and \(\sec x\) in terms of sine and cosine:
\[ \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x \]
This simplifies to:
\[ \frac{\sin x + 1}{\cos x} = 2 \cos x \]

### Step 2: Multiply through by \(\cos x\)
To eliminate the fraction, we multiply both sides by \(\cos x\) (noting that \(\cos x \neq 0\)):
\[ \sin x + 1 = 2 \cos^2 x \]

### Step 3: Use the Pythagorean identity
Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can substitute:
\[ \sin x + 1 = 2(1 - \sin^2 x) \]
Expanding this gives:
\[ \sin x + 1 = 2 - 2\sin^2 x \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ 2\sin^2 x + \sin x - 1 = 0 \]
This is a quadratic equation in terms of \(\sin x\).

### Step 5: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 1, c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ \sin x = \frac{-1 \pm \sqrt{1 + 8}}{4} \]
\[ \sin x = \frac{-1 \pm 3}{4} \]
This gives us two possible solutions:
1. \( \sin x = \frac{2}{4} = \frac{1}{2} \)
2. \( \sin x = \frac{-4}{4} = -1 \)

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

### Step 7: Find the angle for \(\sin x = -1\)
The solution for \( \sin x = -1 \) in the interval \([0, 2\pi]\) is:
\[ x = \frac{3\pi}{2} \]

### Step 8: Verify the solutions
We need to check if any of these solutions make \(\cos x = 0\) (which would make the original equation undefined). The value \(x = \frac{3\pi}{2}\) does make \(\cos x = 0\), so it is not a valid solution.

### Conclusion
The valid solutions are:
1. \( x = \frac{\pi}{6} \)
2. \( x = \frac{5\pi}{6} \)

Thus, the total number of solutions in the interval \([0, 2\pi]\) is **2**.