sin2a+cosa=0?
回答 (2)
2015-08-31 12:15 pm
Suppose that 0 ≤ a ≤ 360°
sin(2a) + cos(a) = 0
2 sin(a) cos(a) + cos(a) = 0
cos(a) [2sin(a) + 1] = 0
cos(a) = 0 or sin(a) = -1/2
[a = 90° or a = 270°] or [a = (180 + 30)° or a = (360 - 30)°]
a = 90°, 210°, 270°, 330°
sin(2a) + cos(a) = 0
2 sin(a) cos(a) + cos(a) = 0
cos(a) [2sin(a) + 1] = 0
cos(a) = 0 or sin(a) = -1/2
[a = 90° or a = 270°] or [a = (180 + 30)° or a = (360 - 30)°]
a = 90°, 210°, 270°, 330°
2015-08-31 12:32 pm
I am very weak of mathematics. it is Trigonometry, I try to solve this but I ma fail. sorry for my weak Knowles.
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