Question

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

Answer 1

To find the number of solutions of the equation tan(x) + sec(x) = 2cos(x) in the interval (0, 2π), we can analyze the behavior of the individual functions involved.

Rewriting the equation using trigonometric identities, we have: sin(x)/cos(x) + 1/cos(x) = 2cos(x)

Combining the fractions on the left-hand side, we get: (sin(x) + 1)/cos(x) = 2cos(x)

Multiplying both sides by cos(x) (assuming cos(x) ≠ 0): sin(x) + 1 = 2cos²(x)

Using the identity sin²(x) + cos²(x) = 1, we substitute cos²(x) = 1 - sin²(x): sin(x) + 1 = 2(1 - sin²(x))

Rearranging the equation, we have: 2sin²(x) + sin(x) - 1 = 0

Factoring the quadratic equation, we get: (2sin(x) - 1)(sin(x) + 1) = 0

This gives us two possible solutions: sin(x) = 1/2 or sin(x) = -1

For sin(x) = 1/2, in the interval (0, 2π), we have x = π/6 and x = 5π/6.

For sin(x) = -1, in the interval (0, 2π), we have x = 3π/2.

Now, we must check for extraneous solutions because we multiplied by cos(x) earlier, which could be zero.

* If x = π/6, tan(π/6) + sec(π/6) = (1/√3) + (2/√3) = 3/√3 = √3. 2cos(π/6) = 2(√3/2) = √3. So, x = π/6 is a solution.
* If x = 5π/6, tan(5π/6) + sec(5π/6) = (-1/√3) + (-2/√3) = -3/√3 = -√3. 2cos(5π/6) = 2(-√3/2) = -√3. So, x = 5π/6 is a solution.
* If x = 3π/2, cos(x) = 0, which means tan(x) and sec(x) are undefined. Therefore, x = 3π/2 is not a solution.

Therefore, the solutions are x = π/6 and x = 5π/6.

Thus, we have found two solutions for the given equation in the interval (0, 2π).