1)Cos2x=1-3cos^2x
2)tg^2x+2tgx=5
Решить уравнение, хелп

1) cos2x = 1 - 3cos²xcos²x - sin²x = 1 - 3cos²xcos²x - 1 + cos²x + 3cos²x = 15cos²x = 2cos²x = 2/5cosx = √0.4cosx = -√0.4x = arccos√0.4 + 2πnx = - arccos√0.4 + 2πnx = arccos(-√0.4) + 2πnx = - arccos(-√0.4) + 2πnn∈Z2) tg²x + 2tgx = 5tg²x + 2tgx - 5 = 0 tgx = tt² + 2t - 5 = 0 t = -1 - √6t = √6 - 1tgx = -1 - √6tgx = √6 - 1x = arctg (-1 - √6) + πnx = arctg (√6 - 1) + πnn∈Z