1) Cos2x=1-3cos^2x

2) tg^2x+2tgx=5

Решить уравнение, хелп

1) cos2x = 1 - 3cos²x

cos²x - sin²x = 1 - 3cos²x

cos²x - 1 + cos²x + 3cos²x = 1

5cos²x = 2

cos²x = 2/5

cosx = √0.4

cosx = - √0.4

x = arccos√0.4 + 2πn

x = - arccos√0.4 + 2πn

x = arccos (-√0.4) + 2πn

x = - arccos (-√0.4) + 2πn

n∈Z

2) tg²x + 2tgx = 5

tg²x + 2tgx - 5 = 0

tgx = t

t² + 2t - 5 = 0

t = - 1 - √6

t = √6 - 1

tgx = - 1 - √6

tgx = √6 - 1

x = arctg (-1 - √6) + πn

x = arctg (√6 - 1) + πn

n∈Z