Question

3-4cos²x=0 сумма корней [0;3п]

Answer 1

4(3/4-cos²x)=0
(√3/2-cosx)(√3/2+cosx)=0
cos x₁=√3/2  x₁=$+- frac{ pi }{6} +2 pi k, k$∈Z
cos x₂=-√3/2  x₂=$+- frac{ 5pi }{6} +2 pi k, k$∈Z
сумма корней = π/6 + 5π/6 + 7π/6+11π/6+(π/6+2π)+ (5π/6+2π)=9π