Solving Trigonometric Equations Using Basic Algebra

The discussion of solving trigonometric equations for this time will use various methods such as those used in algebra. This method involves factoring trigonometric forms, converting equations into quadratic equations, or other methods often used in solving algebraic equations. To solve trigonometric equations this form must also be supported by your ability to apply trigonometric formulas that you have obtained.
Answer:
sin 2x0 + sin x0 = 0
2 sin x0 cos x0 + sin x0 = 0
   sin x0 (2 cos x0 + 1)  = 0
sin x0 = 0 or 2 cos x0 + 1 = 0
(i)            sin x0 = 0 or
(ii)          2 cos x0 = -1/2
(i)            x = 0 + n . 360 or x = 180 + n .360
(ii)          x = 120 + n . 360 or x = -120 + n . 360
By selecting the value of n that corresponds to the interval above, the set of solutions {0, 120, 180, 240, 360}.

Example 1.
Determine the set of completion equations 2 sin2 x0 – 3 sin x0 – 2 = 0, for 0 ≤ x ≤ 360.
Answer:
2 sin2 x0 – 3 sin x0 – 2 = 0

(2 sin2 x0 + 1)( sin x0 – 2) = 0

(i) 2 sin2 x0 + 1 or
(ii) sin x0 – 2


(i) x = 210 + n . 360 or x = - 30 + n . 360
(ii) Do not have a solution because -1 ≤ sin x0 ≤ 1.

By selecting the value of n that corresponds to the interval above, the set of solutions {210, 330}.

Example 2.
Determine the set of completion equations cos 2x0 – cos x0 = 0, for 0 ≤ x ≤ 360.
cos 2x0 – cos x0 = 0
  (2 cos 2 x0 – 1) -  cos x0 = 0
     2 cos 2 x0 – cos x0 - 1 = 0
(2 cos x0 + 1)(cos x0 – 1) = 0

(i) 2 cos x0 + 1 = 0 == > cos x0 = -1/2 
(ii) cos x0 – 1 = 0 == > cos x0 = 1
(i) x = 120 + n . 360 or x = - 120 + n . 360
(ii) x = 0 + n . 360

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

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