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# Solve Trigonometry equations with ax + Trig bx = 0

The discussion of this form includes forms that can be solved using the following formulas,
sin A + sin B = 2 sin ½ (A + B) cos ½ (A – B)
sin A – sin B = 2 cos ½ (A + B) sin ½ (A – B)
cos A + cos B = 2 cos ½ (A + B) cos ½ (A – B)
cos A – cos B = -2 sin ½ (A + B) sin ½ (A – B)

Example 1.
Determine the set of completion equations sin x0 + sin (x – 60)0 = 0, for 0 x 360.
sin x0 + sin (x – 60)0 = 0
2 sin ½ (2x – 60) cos ½ (60)0 = 0
2 sin (x – 30)0 . ½ √3 = 0
√3 sin (x – 30)0 = 0
sin (x – 30)0 = 00
sin (x – 30)0 = sin 00
x – 30 = 0 + n . 360 or x – 30 = 180 + n . 360
x = 30 + n . 360 or x = 210 +  n .  360
by selecting the value n that corresponds to the interval above, then the solution is {30, 210}.

Example 2.
Determine the set of completion equations sin 2x0 + sin x0 = 0, for 0 x 360.
sin 2x0 + sin x0 = 0
2 sin 3/2 x0 cos ½ x0 = 0
(i)            sin 3/2 x0 = 0
sin 3/2 x0 = sin 00
3/2 x = 0 + n . 360 for 3/2 x = 180 + n . 360
x = 0 + n . 240 for x = 120 + n . 240
(ii)          cos ½ x0 = 0
cos ½ x0 = cos 900
½ x = 90 + n . 360 or ½ x = - 90 + n . 360
½ x = 180 + n . 720 or ½ x = - 180 + n . 720

By selecting the value of n that corresponds to the interval above, the set of solutions is {0, 120, 180, 240, 360}.