Question

cos(6x)+6cos^2(3x)=1

Answer 1

cos6x + 6cos²3x = 1
cos²3x - sin²3x + 6cos²3x = 1
7cos²3x - sin²3x = 1
7cos²3x - (1 - cos²3x) = 1
7cos²3x - 1 + cos²3x = 1
8cos²3x = 2
4cos²3x = 1
cos²3x = 1/4

cos3x = 1/2
cos3x = -1/2

3x = -π/3 + 2πn
3x = π/3 + 2πn
3x = 2π/3 + 2πn
3x = -2π/3 + 2πn

x = -π/9 + 2πn/3
x = π/9 + 2πn/3
x = 2π/9 + 2πn/3
x = -2π/9 + 2πn/3
OTBET: -π/9 + 2πn/3; π/9 + 2πn/3; -2π/9 + 2πn/3; 2π/9 + 2πn/3; n ∈ Z