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Solve eqation, where 0°<= x < 360°. Round approximate solutions to the nearest tenth of a degree.

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2sinx - 3cos x = 1?

asked Oct 25, 2014 in ALGEBRA 2 by anonymous

1 Answer

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The trig equation 2sinx - 3cosx = 1

2sinx = 1 + 3cosx

2 sinx - 1 = 3 cosx

2sinx - 1 = 3√(1 - sin2  x)

Apply squre root on each side.

(2sinx - 1)2  = [3√(1 - sin2 x)]2

4sin2x + 1 - 4sinx = 9(1 - sin2 x)

4sin2x + 1 - 4sinx = 9 - 9 sin2 x

4 sin2 x + 9 sin2 x - 4 sinx + 1 - 9 = 0

13 sin2 x - 4 sinx - 8 = 0

Compare it to image

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Roots are image

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Solutions in  0o ≤  x  < 360o

image.

answered Oct 25, 2014 by david Expert

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