Решите уравнения: 1. sin^(2)7x - cos^(2)7x = корень - вопрос №460072112727322 от alena0bashinskaya 28.06.2020 21:42

Question

Алгебра: Решите уравнения: 1. sin^(2)7x - cos^(2)7x = корень из 3 делить на 2. 2. cosx/4 * sinп/5 - sinx/4 * cosп/5 = корень из 3 делить на 2.

alena0bashinskaya · Answer

1. sin² 7x - cos² 7x = √3/2
- (cos² 7x - sin² 7x) = √3/2
- cos (2*7x) = √3/2
- cos 14x = √3/2
cos 14x = -√3/2
14x = + 5π/6 + 2πk, k∈Z
x= + 5π/84 + πk/7, k∈Z

2. cosx/4 * sin π/5 - sinx/4 cos π/5 =√3/2
sin π/5 * cosx/4 - cos π/5 * sinx/4 = √3/2
sin(π/5 - x/4) = √3/2
1) π/5 - x/4 = π/3 +2πn, n∈Z
-x/4 = π/3 - π/5 + 2πn
-x/4 = 2π/15 + 2πn
x= - 8π/15 -8πn, n∈Z
2) π/5 -x/4 = π - π/3 + 2πn,n∈Z
π/5 -x/4 = 2π/3 + 2πn
-x/4 = 2π/3 - π/5 +2πn
-x/4 = 7π/15 + 2πn
x= -28π/15 - 8πn, n∈Z
ответ: -8π/15 - 8πn, n∈Z;
-28π/15 -8πn, n∈Z.