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

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Solve the equation 3 cos x + 2 = sec x for 0 degree _< x _< 360
asked Oct 25, 2014 in TRIGONOMETRY by anonymous

1 Answer

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

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(180o).

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

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

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

If α = 180o then x = 360on ± 180o.

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

If n = 1

x = 360o - 70.53o = 289.47o.

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

 

answered Oct 25, 2014 by casacop Expert
edited Oct 25, 2014 by casacop

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