# 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 x

^{0}+ sin (x – 60)^{0 }= 0, for 0 ≤ x ≤ 360.
Answer:

sin x

^{0}+ 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}= 0^{0}
sin (x – 30)

^{0}= sin 0^{0}
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 2x

^{0}+ sin x^{0 }= 0, for 0 ≤ x ≤ 360.
Answer:

sin 2x

^{0}+ sin x^{0 }= 0
2 sin 3/2 x

^{0}cos ½ x^{0}= 0
(i)
sin 3/2 x

^{0}= 0
sin 3/2 x

^{0}= sin 0^{0}
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 x

^{0}= 0
cos ½ x

^{0}= cos 90^{0}
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}.

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