Question

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

• A. 6
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

2

Answer 2

We can start by simplifying the given equation using the identities: tan(x) = sin(x)/cos(x) sec(x) = 1/cos(x)

Substituting these expressions, we get: sin(x)/cos(x) + 1/cos(x) = 2cos(x)

Multiplying through by cos(x), we get: sin(x) + 1 = 2cos^2(x)

Using the identity sin^2(x) + cos^2(x) = 1, we can write cos^2(x) = 1 - sin^2(x).

Substituting this, we get: sin(x) + 1 = 2(1 - sin^2(x))

Simplifying, we get: 2sin^2(x) + sin(x) - 1 = 0

Using the quadratic formula, we get: sin(x) = (-1 ± √9)/4 sin(x) = -1 or sin(x) = 1/2 For sin(x) = -1, x = 3π/2, which is not in the range (0,2π).

For sin(x) = 1/2, we have x = π/6 and x = 5π/6.

Therefore, the given equation has 2 solutions in the range (0,2π), which is option D.