# 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 2x

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

^{0}cos x^{0}+ sin x^{0 }= 0
sin x

^{0}(2 cos x^{0}+ 1)^{ }= 0
sin x

^{0}= 0 or 2 cos x^{0}+ 1 = 0
(i)
sin x

^{0}= 0 or
(ii)
2 cos x

^{0}= -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 sin

^{2}x^{0}– 3 sin x^{0}– 2 = 0, for 0 ≤ x ≤ 360.
Answer:

2 sin

^{2}x^{0}– 3 sin x^{0}– 2 = 0
(2 sin

^{2}x^{0}+ 1)( sin x^{0}– 2) = 0
(i) 2 sin

^{2}x^{0}+ 1 or
(ii) sin x

^{0}– 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 2x

^{0}– cos x^{0}= 0, for 0 ≤ x ≤ 360.
cos 2x

^{0}– cos x^{0}= 0
(2 cos

^{2}x^{0}– 1) - cos x^{0}= 0
2 cos

^{2}x^{0}– cos x^{0}- 1 = 0
(2 cos x

^{0}+ 1)(cos x^{0}– 1) = 0
(i) 2 cos x

^{0}+ 1 = 0 == > cos x^{0}= -1/2
(ii) cos x

^{0}– 1 = 0 == > cos x^{0}= 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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