6. Решить уравнения, сводящиеся к квадратным: a) 2sin²x = 1; б) 2cos²x + cosx - 3 = 0; в) 3tg²x + tgx - 2 = 0; г) 2cos²x + 3sinx = 0.

a) sin²x = 1/2 => sin x = ±1/√2 => x = ±π/4 + πn, n ∈ Z.
б) Пусть y = cos x. 2y² + y - 3 = 0. D = 1 - 4(2)(-3) = 25. y1 = (-1+5)/4 = 1, y2 = (-1-5)/4 = -3/2. cos x = 1 => x = 2πn, n ∈ Z. cos x = -3/2 (нет решений).
в) Пусть y = tg x. 3y² + y - 2 = 0. D = 1 - 4(3)(-2) = 25. y1 = (-1+5)/6 = 2/3, y2 = (-1-5)/6 = -1. tg x = 2/3 => x = arctg(2/3) + πn, n ∈ Z. tg x = -1 => x = -π/4 + πn, n ∈ Z.
г) 2(1 - sin²x) + 3sinx = 0 => 2 - 2sin²x + 3sinx = 0 => 2sin²x - 3sinx - 2 = 0. Пусть y = sin x. 2y² - 3y - 2 = 0. D = 9 - 4(2)(-2) = 25. y1 = (3+5)/4 = 2, y2 = (3-5)/4 = -1/2. sin x = 2 (нет решений). sin x = -1/2 => x = -π/6 + 2πn, x = 7π/6 + 2πn, n ∈ Z.