cos3x*cosx = cos2x
1/2*[cos(3x-x) + cos(3x + x)] = cos2x
cos2x - cos4x = 2 cos2x
cos4x - cos2x = 0
2*[sin(4x + 2x)/2 * sin(2x - 4x)/2] = 0
1) sin3x = 0
3x = πn, n∈Z
x = πn/3, n∈z
2) sinx = 0
x = πk, k∈Z
Ответ:
б) x²+6x=9
x²+6x-9=0
D=b²-4ac=36-4•1•(-9)=36+36=72
x1= -b+√D/2a= -6+6√2/2=-3+3√2
x2= -b-√D/2a= -6-6√2/2= -3-3√2
a⁻³ * a⁵ = a⁻³⁺⁵ = a²