Question

Solve the equation for x:

sin^2 x – 2 cos x - 2 = 0
for the restriction 0 < x < 2pi

Answer 1

Answer: x = pi

sin²x - 2cos x - 2 = 0

⇔ 1 - cos²x - 2cos x - 2 = 0

⇔ cos²x + 2cos x + 1 = 0

⇔ (cos x + 1)² = 0

⇒ cos x + 1 = 0

⇔ cos x = -1

⇒ x = pi + k2pi

because 0 ≤ x < 2pi

=> x = pi

Explanation: