В) <span>.
2*{[cos(x + 3x)/2] * [cos(x - 3x)/2]}= cos2x
2*[cos2x*cosx] - cos2x = 0
cos2x(2cosx - 1) = 0
1) cos2x = 0
2x = </span>π/2 + πn, n∈Z
<span>x1 = </span>π/4 + (πn)/2, n ∈Z
<span>2) 2cosx - 1 = 0
cosx = 1/2
x = (+ -) arccos(1/2) + 2</span>πk, k∈Z
<span>x2 = (+ -)*(</span>π/3) + 2πk, k∈Z <span>
</span>
( x + 7) /3 = ( 2x + 3) / 5
5( x + 7) = 3 ( 2x + 3)
5x + 35 = 6x + 9
5x - 6x = - 35 + 9
- x = - 26
x = 26
Y ' = (x^2 - 6*x + 1) ' = 2*x - 6
y ' (-1) = 2*(-1) - 6 = - 2 - 6 = - 8