A) 4cos^4x-3cos2x-1=0

b) найдите все корни уравнения принадлежащие интервалу (-7π:2; -2π)

4cos^4x-3cos2x-1=0

4 * (1+cos2x) ²/4 - 3cos2x-1=0

1+2cos2x+cos²2x-3cosx-1=0

cos²2x-cosx=0

cos2x (cos2x-1) = 0

cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2

cos2x=1⇒2x=2πn⇒x=πn

x={-13π/4; -11π/4; -9π/4}∈ (-7π:2; -2π)