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