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## Sove a trigonometry equation

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

- Solution:
- 2 sin x + 2 cos x = (square root of 3) - 1
- Both side of the equation divide by 2,
- sin x + cos x = (square root of 3 - 1)/2
- square both side of the equation,
- (sin x + cos x)
^{2}= [(square root of 3 -1)/2]^{2} - apply the formula, (a + b)
^{2}= a^{2}+ 2ab + b^{2}and (c/d)^{2}= c^{2}/d^{2}, - sin
^{2}x + 2sin x cos x + cos^{2}x = [(square root of 3)^{2}- 2(square root of 3) + 1^{2}]/4 = [3 - 2(square root of 3 + 1)]/4 = 4/4 - 2(square root of 3)/4 = 1 - (square root of 3)/2 - Apply the formula, sin
^{2}x + cos^{2}x = 1, - 1 + 2sin x cos x = 1 - (square root of 3)/2
- Bothe sides of the equation subtract 1,
- 2sin x cos x = -(square root of 3)/2
- apply the formula, sin 2x = 2 sin x cos x
- sin 2x = -(square root of 3)/2
- Because sin 2x < 0 and sin60
^{o}= (square root of 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 = 180^{o} + 60^{o} = 240^{o}, so, x = 120^{o}, then, x = k × 360^{o} + 120^{o}, in which, k = 0,1,2,3...

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