Решите, ! sinb*cosb+2/5cos^2+1,если tgb=2. sinb*cosb-3/6cos^2b-sin^2b, если tgb=-2. 2sinb*cosb+3/4cos^2b+sin^2b,если tgb-4. cos^2b+2/cos^2b+3sinb*cosb,если tgb=3.

Объяснение:

https://tex.z-dn.net/?f=%28x-1%29%28x%5E2%2B4x%2B4%29%3D4%20%28x%2B2%29%20%5C%5C%28x-1%29%28x%2B2%29%5E2-4%28x%2B2%29%3D0%5C%5C%28x%2B2%29%28%28x-1%29%28x%2B2%29-4%29%3D0%5C%5Cx%2B2%3D0%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%28x-1%29%28x%2B2%29-4%3D0%5C%5Cx_1%3D-2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20x%5E2%2Bx-6%3D0%5C%5C.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20x_%7B2%2C3%7D%3D%5Cfrac%7B-1%5E%2B_-%5Csqrt%7B1%2B24%7D%7D%7B2%7D%3D%5Cfrac%7B-1%5E%2B_-5%7D%7B2%7D%5C%5C.%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20x_2%3D-3%5C%20%5C%20%5C%20%5C%20x_3%3D2%5C%5COTBET%3Ax_1%3D-2%5C%20x_2%3D-3%5C%20x_3%3D2

2cos(π/3 - 3x) + √3 = 0

2cos(π/3 - 3x) = -√3

cos(π/3 - 3x) = -√3/2

• Воспользуемся формулой:

cos(x) = b ( |b|≤ 1, [0; π] )

x = ± arccos(b) + 2πn, n ∈ ℤ

• Получаем:

cos(π/3 - 3x) = -√3/2

π/3 - 3x = ± arccos(-√3/2) + 2πn, n ∈ ℤ

π/3 - 3x = ± (π - arccos(-√3/2)) + 2πn, n ∈ ℤ

π/3 - 3x = ± (π - 5π/6) + 2πn, n ∈ ℤ

π/3 - 3x = ± π/6 + 2πn, n ∈ ℤ

-3x = ± π/6 - π/3 + 2πn, n ∈ ℤ

[ -3x = -π/6 - π/3 + 2πn, n ∈ ℤ

[ -3x = π/6 - π/3 + 2πn, n ∈ ℤ

[ -3x = -π/2 + 2πn, n ∈ ℤ / : (-3)

[ -3x = -π/3 + 2πn, n ∈ ℤ / : (-3)

[ x = π/6 - 2πn/3, n ∈ ℤ

[ x = π/9 - 2πn/3, n ∈ ℤ

ответ: x = π/6 - 2πn/3, n ∈ ℤ ; x = π/9 - 2πn/3, n ∈ ℤ