數學問題,關於sin,cos,tan,我諗極都唔識做..."

1a)
(sin225 cos315 - tan120)/[sin(-60)tan210]

= [sin(180+45)cos(360-45)-tan(180-60)]/[-sin60tan(180+30)]
= [-sin45cos45-(-tan60)]/[(-sin60)(tan30)]
= [-(sq root 2)/2 * (sq root 2)/2 + sq root 3]/[-(sq root 3)/2/(sq root 3)]
= (sq root 3 - 1/2)*(-2)
= 1 - 2(sq root 3)

b) 5sin217 cos233 + 5sin127 cos323
= 5[sin(180+37)cos(270-37)+sin(90+37)cos(270+37)]
= 5[(-sin37)(-sin37)+(cos37)(cos37)]
= 5[sin^2(37)+cos^2(37)]
= 5

3)
sin 2X=0.5
2X = 360n+30 or 2X = 360n +150
X = 15, 75, 195, 255. Therefore, there are 4 roots between 0 and 360

4.
5sin^2(X) - 2sinXcosX - 3cos^2(X) = 0
(5sinX + 3cosX)(sinX - cosX) = 0
5sinX + 3cosX = 0 or sinX - cosX = 0
tanX = -5/3 or tanX = 1
X = 120.96 or 300.96 or 45 or 225

5.Solve the equation 4sinA=3tanA for 0<=A<=360
4sinA = 3tanA
4tanAcosA = 3tanA
4tanAcosA - 3tanA = 0
tanA(4cosA - 3) = 0
tanA = 0 or cosA = 3/4
A = 0 or 180 or 41.41or 318.59

6
cos(3X - 45)= 1/2
3X-45 = 360n + 60 or 3X-45 = 360n - 60
X = 35 or 155 or 275 or 115 or 235 or 355