How do I solve this trig equation 40=20sin(x)+40cos(x)sin60 ? the angles are in degress?

40 = 20 sin(x) + 40 cos(x) sin60

Divide both sides by 80 :
1 = (1/2) sin(x) + cos(x) sin60°
Thus, (1/2) sin(x) + cos(x) sin60° = 1

cos60° = 1/2, the equation becomes :
cos(60°) sin(x) + cos(x) sin(60°) = 1

Apply the identity sinA cosB + cosB sinA = sin(A + B) :
sin(60° + x) = 1

In the range of 0° ≤ x ≤ 360°, only sin(90°) = 1 :
sin(60° + x) = sin(90°)
60° + x = 90°
x = 30°

2=sinx+sqrt(3)cosx
2secx = tanx+sqrt(3)
4sec^2(x) = tan^2(x)+2sqrt(3)tanx+3
4[1+tan^2(x)] = tan^2(x)+2sqrt(3)tanx+3
3tan^2(x)]-2sqrt(3)tanx+1 = 0
(sqrt(3)tanx-1)^2 = 0
tanx = 1/sqrt(3)
x = 30 degrees

you really have 1 = [ 1 / 2] sin x + [ √3 / 2 ] cos x ≡ sin ( x + 60) ---> x = 30°