solve in degrees
1) 2sin2 (x/4) = 1
2) 2 cos (3x) + 1 = 0
3) cot2 (x + 18°) = 1
4) (1 + tan(x + 87°)) / 2 = 7
Usually there is a domain given for these kind of question, I will assum 0 ≤ x ≤ 360°
Here is how I do these kind of trig equations.
I will do 2nd question, you try the others in the same way
1. simply the trig function as far as possible
2 cos(3x) + 1 = 0
cos(3x) = -1/2
2. Using the CAST rule establish in which quadrant your angle lies
since the cos(3x) is negative, 3x must fall in quad II or III
3. find the reference angle by using the positive only, that is,
find Ø so that cosØ = +1/2
so the reference angle is 60°
3x = 180-60 = 120°
x = 40°
or
3x = 180 + 60 = 240°
x = 80°
4. since the period of cos(3x) is 360/3 or 120°
we can obtain more solutions by adding or subtracting 120° to any previous solution until we run beyond the given domain.
so other solutions:
x = 40+120 = 160°
x = 160+120 = 280°
x = 280+120 = 400° ---> too big
x = 80+120 = 200°
x = 200+120 = 320°
so we have x = 40, 80, 160, 200, 280, 320
(use your calculator to test them, they all work)