Решение
<span>sin²x/4-cos²x/4=1/2
- (cos</span>²x/4 - sin²x/4) = 1/2
cos<span>²x/4 - sin²x/4 = - 1/2
</span>cos[2*(x/4)] = - 1/2
cosx/2 = - 1/2
x/2 = +-arccos(-1/2) + 2πk, k ∈ Z
x/2 = +- [π - arccos(1/2)] + 2πk, k ∈ Z
x/2 = +- [π - π/3<span>)] + 2πk, k ∈ Z
</span>x/2 = +- [2π/3<span>)] + 2πk, k ∈ Z
</span>x = +- [4π/3<span>)] + 4πk, k ∈ Z</span>
1)2х=27+5(переносим с противоположным знаком!!! )2х=32 х=32:2 х=16 2)4у=-5+3 4у=-2 у=-2:4 у=-0.5. 3)2х-4х=3+1 -2х=4 х=4:(-2) х=-2