Question

Prosze o pomoc! Rozwiazac rownanie

1)2sin²x = cos2x (π≤x≤3π/2)

2)2cos²x + cos2x = 0 ( -π≤x≤-π/2)

Dziekuje!

Answer 1

2cos^2(x) = 3sin x + 3

2(1 - sin^2(x)) = 3sin x + 3

2 - 2sin^2(x) = 3sin x + 3

2sin^2(x) + 3sin x + 1 = 0

(2sin x + 1)(sin x + 1) = 0

case 1: 2sin x + 1 = 0

sin x = -1/2

x = 7Π /6, 11Π/6

case 2: sin x + 1 = 0

sin x = -1

x = 3Π/2

x = 7Π/6, 3Π/2 and 11Π/6 in the interval [ 0, 2Π ]