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How to solve Sin2x + 3Cos2x = 0 ?

0 votes
0<x<180
asked Apr 12, 2013 in TRIGONOMETRY by homeworkhelp Mentor

1 Answer

0 votes

0 < x < π

sin2x + 3cos2x = 0

Subtract 3cos2x from each side

sin2x  + 3cos2x - 3cos2x = -3cos2x

sin2x + 0 = -3cos2x

sin2x = -3cos2x

Divide cos2x by each side

sin2x / cos2x = -3cos2x / cos2x

tan2x = -3

Apply arctan each side

arctan(tan2x) = arctan(-3)

Simplify

2x = arctan(-3)

Divide each side by 2

x = 1 / 2 arctan(-3).

 

 

 

answered Apr 12, 2013 by diane Scholar

tan (2x) = - 3.

tan (2x) = tan(- 71.560).

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

2x = nπ - 71.56

⇒ x = nπ/2 - 35.78

If, n = 0, x = (0)π/2 - 35.78 = - 35.78,

If n = 1, x = (1)π/2 - 35.78 = 90 - 35.78 = 54.22,

If n = 3, x = (2)π/2 - 35.78 = 180 - 35.78 = 144.22.

Therefore, the solutions are x = 54.220 and x = 144.220 in the interval (0, π).

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