Question

1)4cos^2x=3
2)4sin^2x-cosx-1=0

Answer 1

1)4cos²x = 3

2cos²x = 3/2

1+cos2x = 3/2

cos2x = 1/2

2x = ±π/3 + 2πk

x = ±π/6 + πk

2)4sin²x - cosx - 1 = 0

4(1-cos²x) - cos x - 1 = 0

4-4cos²x - cosx - 1 = 0

4cos²x + cosx - 3 = 0

cos x = a

4a²+a-3= 0

a1 = -1

a2 = 3/4

a)cos x = -1

x = π + 2πk

б)cos x = 3/4

x = ±arccos(3/4) + 2πk