2cos²x = 1 + sinx
2(1 - sin²x) = 1 + sinx
2 - 2sin²x = 1 + sinx
2sin²x + sinx - 1 = 0
Пусть sinx = t, причем t ∈ [-1;1]
2t² + t - 1 = 0
D = 1 + 8 = 9
t = ( - 1 + 3)/4 = 1/2;
t = ( - 1 - 3)/4 = - 1;
1) sinx = 1/2
x = (-1)^k*arcsin(1/2) + πk
x = (-1)^k*π/6 + πk, k ∈ Z
2) sinx = -1
x = - π/2 + 2πk, k ∈ Z
Ответ:
(-1)^k* π/6 + πk;
-π/2 + 2πk, k ∈ Z
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