sin(θ-30°)=2cos(θ+60°)

sin(θ-30°)=2cos(θ+60°)
sin θ cos 30* - sin 30* cos θ = 2 cos θ cos 60* - 2 sin θ sin 60*
√3/2 sin θ - 1/2 cos θ = 2(1/2) cos θ - 2 (√3/2 ) sin θ
√3/2 sin θ - 1/2 cos θ = cos θ - √3 sin θ
3√3/2 sin θ = 3 / 2 cos θ
√3 sin θ = cos θ
tan θ = 1/ √3
θ = 180°n+30°where n is an integer

sin(θ-30°)=2cos(θ+60°)
cos[90°-θ+30°]=2cos(θ+60°)
cos(θ-120°)=2cos(θ+60°)
-cos[180+θ-120°]=2cos(θ+60°)
-cos(θ+60°)=2cos(θ+60°)
3cos(θ+60°)=0
cos(θ+60°)=0
θ+60°=360°n±90°
θ=360°n+30° or θ=360°n-150° where n is an integer.

they are correct,but you can simplify it！！

let n be 1,
θ=360°n+30° or θ=360°n-150°
θ=360°(1)+30° or θ=360°(1)-150°
θ=390° or θ=210°

which is 180°(2)+30° or 180°(1)+30°

Since n can be any integer,the general form will be θ=180°n+30° where n is an integer.

you can also draw graphs yourself,then you will understand it!!!!!!!!!!!!!!!!!!!

sin(θ-30°)=2cos(θ+60°)
sin[-90+(θ+60)]=2cos(θ+60)
-sin[90-(θ+60)]=2cos(θ+60)
-cos(θ+60)=2cos(θ+60)
3cos(θ+60)=0
cos(θ+60)=cos(90)
cosθ=cos30<------HERE!!!
θ=180n+30//

θ=360°n+30° = 180°(2n)+30°
θ=360°n-150° = 360°n - 180° +30° = 180°(2n-1) + 30°
Hence, you are correct but you represent it in a different way.

2007-06-15 18:38:01 補充：
Do you notice that 180°(2n) + 30° means 180°(even integers) + 30° and that 180°(2n - 1) + 30° means 180°(odd integers) + 30°? Hence this means you have already exhausted all the answers for 180°n + 30°, which means 180°(integers) + 30°.