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 = a² + 2ab + b² and (c/d)² = c²/d²,

sin²x + 2sin x cos x + cos²x = [(square root of 3)² - 2(square root of 3) + 1²]/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, sin²x + cos²x = 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...