Q:

Solve the equation 2cos^2x + 3sinx = 3

Accepted Solution

A:
Answer:x = 2 Ο€ n_1 + Ο€/2 for n_1 element Z or x = 2 Ο€ n_2 + (5 Ο€)/6 for n_2 element Z or x = 2 Ο€ n_3 + Ο€/6 for n_3 element ZStep-by-step explanation:Solve for x: 2 cos^2(x) + 3 sin(x) = 3 Write 2 cos^2(x) + 3 sin(x) = 3 in terms of sin(x) using the identity cos^2(x) = 1 - sin^2(x): -1 + 3 sin(x) - 2 sin^2(x) = 0 The left hand side factors into a product with three terms: -(sin(x) - 1) (2 sin(x) - 1) = 0 Multiply both sides by -1: (sin(x) - 1) (2 sin(x) - 1) = 0 Split into two equations: sin(x) - 1 = 0 or 2 sin(x) - 1 = 0 Add 1 to both sides: sin(x) = 1 or 2 sin(x) - 1 = 0 Take the inverse sine of both sides: x = 2 Ο€ n_1 + Ο€/2 for n_1 element Z or 2 sin(x) - 1 = 0 Add 1 to both sides: x = 2 Ο€ n_1 + Ο€/2 for n_1 element Z or 2 sin(x) = 1 Divide both sides by 2: x = 2 Ο€ n_1 + Ο€/2 for n_1 element Z or sin(x) = 1/2 Take the inverse sine of both sides: Answer: x = 2 Ο€ n_1 + Ο€/2 for n_1 element Z or x = 2 Ο€ n_2 + (5 Ο€)/6 for n_2 element Z or x = 2 Ο€ n_3 + Ο€/6 for n_3 element Z