Cos2x=sinx+1 .........

Решение Cos2x=sinx+1 1 - 2sinx - sinx = 1 2sinx + sinx = 0 sinx(2sinx + 1) = 0 1) sinx = 0 x = k, k Z 2) 2sinx + 1 = 0 2sinx = - 1 sinx = - 1/2 x = (-1)^n*arcsin(-1/2) + n, n Z но arcsin (- 1/2) = - arcsin (1/2) = - /6, тогда х= (-1) /6 +n, nZ