Question

What is the number of solutions of tanx + secx = 2 cosx if x belongs to (0, 2π)?

Answer 1

We are tasked with finding the number of solutions to the equation \[ \tan(x) + \sec(x) = 2\cos(x) \] in the interval \( (0, 2\pi) \). Let's analyze the behavior of the functions involved.

Step 1: Rewrite the equation using trigonometric identities:
The tangent and secant functions can be rewritten as: \[ \frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)} = 2\cos(x) \]

Step 2: Combine the fractions on the left-hand side:
To combine the fractions, we get: \[ \frac{\sin(x) + 1}{\cos(x)} = 2\cos(x) \]

Step 3: Multiply both sides by \( \cos(x) \):
Assuming \( \cos(x) \neq 0 \), we multiply both sides by \( \cos(x) \) to eliminate the denominator: \[ \sin(x) + 1 = 2\cos^2(x) \]

Step 4: Use the Pythagorean identity:
Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), we substitute \( \cos^2(x) = 1 - \sin^2(x) \) into the equation: \[ \sin(x) + 1 = 2(1 - \sin^2(x)) \]

Step 5: Rearrange the equation:
Expanding and simplifying: \[ \sin(x) + 1 = 2 - 2\sin^2(x) \] 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 6: Factor the quadratic equation:
We can factor the quadratic equation as: \[ (2\sin(x) - 1)(\sin(x) + 1) = 0 \] This gives two possible solutions: \[ \sin(x) = \frac{1}{2} \quad \text{or} \quad \sin(x) = -1 \]

Step 7: Solve for \( x \):
- For \( \sin(x) = \frac{1}{2} \), in the interval \( (0, 2\pi) \), the solutions are: \[ x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6} \] - For \( \sin(x) = -1 \), in the interval \( (0, 2\pi) \), the solution is: \[ x = \frac{3\pi}{2} \]

Step 8: Check for extraneous solutions:
Since we multiplied both sides of the equation by \( \cos(x) \), we need to check if any solution results in \( \cos(x) = 0 \), which would make \( \tan(x) \) and \( \sec(x) \) undefined.
- For \( x = \frac{\pi}{6} \), we calculate: \[ \tan\left( \frac{\pi}{6} \right) + \sec\left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \] and \[ 2\cos\left( \frac{\pi}{6} \right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \] Thus, \( x = \frac{\pi}{6} \) is a valid solution.
- For \( x = \frac{5\pi}{6} \), we calculate: \[ \tan\left( \frac{5\pi}{6} \right) + \sec\left( \frac{5\pi}{6} \right) = \frac{-1}{\sqrt{3}} + \frac{-2}{\sqrt{3}} = \frac{-3}{\sqrt{3}} = -\sqrt{3} \] and \[ 2\cos\left( \frac{5\pi}{6} \right) = 2 \times \frac{-\sqrt{3}}{2} = -\sqrt{3} \] Thus, \( x = \frac{5\pi}{6} \) is a valid solution.
- For \( x = \frac{3\pi}{2} \), \( \cos(x) = 0 \), so \( \tan(x) \) and \( \sec(x) \) are undefined. Therefore, \( x = \frac{3\pi}{2} \) is not a solution.

Final Answer:
Therefore, the solutions to the equation in the interval \( (0, 2\pi) \) are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).