solve for x if 2cos(x-30)=1, £(-90;270)

Answer 1

don't know where to start

Answer 2

recall:
cos(A-B) = cosAcosB + sinAsinB

so ....
2cos(x-30) = 1
2(cosxcos30 + sinxsin30) = 1
cosx(√3/2) + sinx (1/2) = 1/2
times 3
√3cosx + sinx = 1
√3cosx = 1 - sinx
square both sides
3cos^2 x = 1 - 2sinx + sin^2 x
3(1 - sin^2 x) = 1 - 2sinx + sin^2 x
...
4sin^2 x - 2sinx -2 = 0
2sin^2 x - sinx - 1 = 0
(2sinx + 1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
x = 210° or 330° or x = 90°

since we squared, all answers should be tested in the original question

if x = 210,
LS = 2cos(180) = -2 ≠ 1
if x = 330,
LS = 2cos300 = 1 = RS, so x = 330°
if x = 90
LS = 2cos(60) = 1 = RS , so x = 90°

x = 90° or x = 330°
in radians:
x = π/2, 11π/6