Решите тригонометрическое уравнение:

1) cos2x+2sin2x+2=0

2) 2cos2x+2sin^2x=5+4sin2x

3) 2sin2x=3-2sin^2x

1) cos2x+2sin2x+2=0

(1 - tg²x) / (1 + tg²x) + 2*2tgx / (1 + tg²x) + 2 = 0 | * (1 + tg²x) ≠ 0

1 - tg²x + 4tgx + 2 (1 + tg²x) = 0

1 - tg²x + 4tgx + 2 + 2tg²x = 0

tg²x + 4tgx + 3 = 0

По т. Виета корни - 1 и - 3

а) tgx = - 1 б) tgx = - 3

x = - π/4 + πk, k ∈Z x = - arctg 3 + πn, n ∈Z

2) 2cos2x+2sin²x=5+4sin2x

2 (Cos²x - Sin²x) + 2Sin²x = 5 + 8SinxCosx

2Cos²x - 2Sin²x + 2Sin²x = 5*1 + 8SinxCosx

2Cos²x = 5 * (Sin²x + Cos²x) + 10SinxCosx

2Cos²x = 5Sin²x + 5Cos²x + 8SinxCosx

5Sin²x + 3Cos²x + 8sinxCosx = 0 |: Cos²x

5tg²x + 8tgx + 3 = 0

tgx = t

5t² + 8t + 3 = 0

t = (-4 + - √ (16 - 15)) / 5 = (-5 + - 1) / 5

t₁ = - 6/5 = - 1.2 t₂ = - 4/5 = - 0,8

tgx = - 1,2 tgx = - 0,8

x = - arctg1,2 + πk, k ∈ Z x = - arctg0,8 + πn, n ∈Z

3) 2sin2x=3-2sin²x

6SinxCosx = 3Sin²x + 3Cos²x - 2Sin²x

6SinxCosx = Sin²x + 3Cos²x | : Cos²x

6tgx = tg²x + 3

tgx = t

t² - 6t + 3 = 0

t = 3+-√ (9 - 3)

t₁ = 3 + √6 t₂ = 3 - √6

tgx = 3 + √6 tgx = 3 - √6

x = arctg (3 + √6) + πk, k ∈Z x = arctg (3 - √6) + πn, n ∈ Z