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## Sove a trigonometry equation

Question: 2 sin x + 2 cos x = Sqrt (3) - 1, find the value of x in degree measure.

Solution:
2 sin x + 2 cos x = Sqrt (3) - 1
Both side of the equation divide by 2,
sin x + cos x = [Sqrt (3) - 1]/2
square both side of the equation,
(sin x + cos x)2 = [(Sqrt (3) -1)/2]2
apply the formula, (a + b)2 = a2 + 2ab + b2 and (c/d)2 = c2/d2,
sin2x + 2sin x cos x + cos2x
= [Sqrt (3)2 - 2 (Sqrt (3) + 12]/4
= [3 - 2 Sqrt (3) + 1]/4
= 4/4 - 2 Sqrt (3)/4
= 1 - Sqrt (3)/2
Apply the formula, sin2x + cos2x = 1,
1 + 2 sin x cos x = 1 - Sqrt (3)/2
Bothe sides of the equation subtract 1,
2 sin x cos x = -Sqrt (3)/2
apply the formula, sin 2x = 2 sin x cos x
sin 2x = -Sqrt (3)/2
Because sin 2x < 0 and sin60o = Sqrt (3)/2,
so, the angle 2x should be in quadratic III ot quadratic IV.

If the terminal side of the angle 2x lies in quadratic III, then, 2x = 180o + 60o = 240o, so, x = 120o, then, x = k × 360o + 120o, in which, k = 0,1,2,3...

If the terminal side of the angle 2x lies in quadrant IV, then 2x = -60o, x = -30o, so x = k × 360o - 30o,  in which, k = 0,1,2,3...