解下列方程

1.
0° ≤ x ≤ 360°
0° ≤ 2x ≤ 720°

2sin2x = -√3
sin2x = -(√3)/2
2x = (180 + 60)°, (360 - 60)°, (540 + 60)°, (720 - 60)°
x = 120°, 150°, 300°, 330°


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2.
0° ≤ x ≤ 360°
-15°≤ x - 15°≤ 345°

3tan(x - 15°) + √6 = 0
3tan(x - 15°) = -√6
tan(x - 15°) = -(√6)/3
(x - 15°) = (180 - 39.23)°, (360 - 39.23)°
x = 155.77°, 335.77


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3.
sin²x - (√3)sinx = 0
sinx(sinx - √3) = 0
sinx = 0 或 sinx = √3 (捨去)
sinx = 0°, 360°

（若題目真的是 0° < X < 360°，則此題無解。）


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4.
3tan²x - tanx - 4 = 0
(3tanx - 4)(tanx + 1) = 0
tanx = 4/3, -1
x = 53.13°, (180 + 53.13)°, (180 - 45)°, (360 - 45)°
x = 53.13°, 233.13°, 135°, 315°

2012-03-19 02:52:22 補充：
5.
sin²x - 3cosx + 3 = 0
(1 - cos²x) - 3cosx + 3 = 0
-cos²x - 3cosx + 4 = 0
cos²x + 3cosx - 4 = 0
(cosx - 1)(cosx + 4) = 0

若 0° ≤ x ≤ 360°
則cosx = 1 或 cosx = -4(捨去)
x = 0°, 360° ...... (答案)

若 0° < x < 360°
則cosx = 1(捨去) 或 cosx = -4(捨去)
故此題無解。 ...... (答案)

不好意思，SIN，COS，等等的完全不會。