Question

Solve the equation 3 cos x + 2 = sec x for 0 degree _< x _< 360

Answer 1

The equation is 3 cos(x) + 2 = sec(x) and interval is 0^o ≤ x ≤ 360^o.

3 cos(x) + 2 = 1/cos(x)

3 cos²(x) + 2 cos(x) - 1 = 0

3 cos²(x) + 3 cos(x) - cos(x) - 1 = 0

3 cos(x)[cos(x) + 1] -1[cos(x) + 1] = 0

[3 cos(x) -1] [cos(x) + 1] = 0

cos(x) = 1/3 and cos(x) = -1.

cos(x) = cos[cos^-1(1/3)] and cos(x) = cos(180^o).

cos(x) = cos(70.53^o) and cos(x) = cos(180^o).

General solution: θ = 360^on ± α, where n is an integer.

If α = 70.53^o then x = 360^on ± 70.53^o.

If α = 180^o then x = 360^on ± 180^o.

If n = 0 then x = 70.53^o and 180^o.

If n = 1

x = 360^o - 70.53^o = 289.47^o.

The solutions are x = 70.53^o, x = 180^o and x = 289.47^o.