Find all solutions of the equation in the interval (0, 2π). Write your answer in radians in terms of π. -5cosx=-2sin^2 x+4

Answer 1

Rewriting the given equation,

2sin²(x) - 5cos(x) - 4 = 0

Using the identity sin²(x) = 1 - cos²(x) and substituting,

2 - 2cos²(x) - 5cos(x) - 4 = 0

-2cos²(x) - 5cos(x) - 2 = 0

Solving for cos(x) using the quadratic formula,

cos(x) = (5 ± √(5² - 4*(-2)*(-2)))/(-4)

cos(x) = (5 ± √9)/(-4)

cos(x) = (5 ± 3)/(-4)

We must take the solution using the minus sign, otherwise we would have cos(x) < -1 which cannot be. Thus,

cos(x) = (5 - 3)/(-4) = (-2)/(-4) = 1/2

We are looking for every x value such that x ∈ (0, 2π) and cos(x) = 1/2. We know that cos(x) is positive in quadrants 1 and 4, and that cos(π/3) = 1/2. Thus, one solution is x = π/3 and the other is the representative angle in quadrant 4 which is x = 5π/3.