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