Question

The number of solutions of tanx+secx=2cosx, n(0,2π) are?

• A. 6
• B. 4
• C. 3
• D. 2

Answer 1

The Correct Option is D

We are tasked with solving the trigonometric equation. Let's break it down step by step.

Step 1: Start with the given equation:
The given equation is: \[ \frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)} = 2 \cos(x) \]

Step 2: Multiply through by \( \cos(x) \):
To eliminate the denominator, we multiply both sides of the equation by \( \cos(x) \): \[ \sin(x) + 1 = 2 \cos^2(x) \]

Step 3: Use the identity \( \sin^2(x) + \cos^2(x) = 1 \):
We use the Pythagorean identity to express \( \cos^2(x) \) in terms of \( \sin^2(x) \): \[ \cos^2(x) = 1 - \sin^2(x) \] Substituting this into the equation, we get: \[ \sin(x) + 1 = 2(1 - \sin^2(x)) \]

Step 4: Simplify the equation:
Expanding the right side: \[ \sin(x) + 1 = 2 - 2 \sin^2(x) \] Bringing all terms to one side: \[ 2 \sin^2(x) + \sin(x) - 1 = 0 \] This is a quadratic equation in terms of \( \sin(x) \).

Step 5: Solve using the quadratic formula:
The quadratic equation is: \[ 2 \sin^2(x) + \sin(x) - 1 = 0 \] Applying the quadratic formula: \[ \sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For \( a = 2, b = 1, c = -1 \), we get: \[ \sin(x) = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} \] Simplifying: \[ \sin(x) = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} \] \[ \sin(x) = \frac{-1 \pm 3}{4} \] So we get two possible solutions: \[ \sin(x) = \frac{-1 + 3}{4} = \frac{1}{2} \quad \text{or} \quad \sin(x) = \frac{-1 - 3}{4} = -1 \]

Step 6: Find the values of \( x \):
For \( \sin(x) = -1 \), we have: \[ x = \frac{3\pi}{2}, \quad \text{which is not in the range } (0, 2\pi) \] For \( \sin(x) = \frac{1}{2} \), we have: \[ x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6} \] These are the two solutions in the range \( (0, 2\pi) \).

Final Answer:
Therefore, the given equation has two solutions in the range \( (0, 2\pi) \), which corresponds to option (D).