Number of solutions of the equation tan x + sec $x=2~cos~x$, lying in the interval [0, 2π] is :

Question

Accepted Answer

To determine the number of solutions for the equation $\tan x + \sec x = 2 \cos x$ in the interval $[0, 2\pi]$, we need to analyze the behavior of the trigonometric functions involved.

1. **Rewriting the Equation:**
   The given equation is $\tan x + \sec x = 2 \cos x$.
   - Recall that $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$.
   - Substitute these into the equation: $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$.
   - Combine the terms on the left: $\frac{\sin x + 1}{\cos x} = 2 \cos x$.
   - Multiply both sides by $\cos x$ (assuming $\cos x \neq 0$): $\sin x + 1 = 2 \cos^2 x$.

2. **Using Trigonometric Identities:**
   - Recall the identity $\cos^2 x = 1 - \sin^2 x$.
   - Substitute this into the equation: $\sin x + 1 = 2(1 - \sin^2 x)$.
   - Simplify: $\sin x + 1 = 2 - 2 \sin^2 x$.
   - Rearrange: $2 \sin^2 x + \sin x - 1 = 0$.

3. **Solving the Quadratic Equation:**
   - This is a quadratic equation in terms of $\sin x$: $2 \sin^2 x + \sin x - 1 = 0$.
   - Use the quadratic formula: $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 1$, and $c = -1$.
   - Calculate the discriminant: $b^2 - 4ac = 1 + 8 = 9$.
   - The solutions for $\sin x$ are: $\sin x = \frac{-1 \pm 3}{4}$.
   - This gives $\sin x = \frac{1}{2}$ and $\sin x = -1$.

4. **Finding Solutions in the Interval [0, 2π]:**
   - 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 in $[0, 2\pi]$ is $x = \frac{3\pi}{2}$.

5. **Verification:**
   - Verify each solution in the original equation to ensure they satisfy $\tan x + \sec x = 2 \cos x$.
   - Upon verification, it is found that only two of these solutions satisfy the original equation.

Therefore, the number of solutions of the equation $\tan x + \sec x = 2 \cos x$ in the interval $[0, 2\pi]$ is indeed two, which supports Option 2 as the correct answer.

**Why Other Options are Incorrect:**
- **Option 1 (One solution):** This is incorrect because we found two valid solutions that satisfy the equation.
- **Option 3 (Three solutions):** This is incorrect because only two solutions satisfy the equation upon verification.
- **Option 4 (Four solutions):** This is incorrect because there are only two solutions that satisfy the equation in the given interval.

Thus, Option 2 is the correct answer, as there are exactly two solutions in the interval $[0, 2\pi]$.