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

Question

2sinx - 3cos x = 1?

Answer 1

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

Roots are

Solutions in  0^o ≤  x  < 360^o

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