更新1:
Solve each equation for solutions over the interval [0º, 360º) (nearest tenth) and radians [0, 2π) (2 decimals). (least possible non-negative angles.)
Trig practice help please. sin x/2 = √(2) - sin x/2?
回答 (2)
2016-04-26 4:56 am
For 0° ≤ x ≤ 360°
Then 0° ≤ x/2 ≤ 180°
sin(x/2) = (√2) - sin(x/2)
2 sin(x/2) = (√2)
sin(x/2) = (√2)/2
x/2 = 45° or (180°-45°)
x/2 = 45° or 135°
x = 90° or 270°
Express in radians :
x = π/2 or 3π/2
x = 1.57 or 4.71 (2 decimal places)
Then 0° ≤ x/2 ≤ 180°
sin(x/2) = (√2) - sin(x/2)
2 sin(x/2) = (√2)
sin(x/2) = (√2)/2
x/2 = 45° or (180°-45°)
x/2 = 45° or 135°
x = 90° or 270°
Express in radians :
x = π/2 or 3π/2
x = 1.57 or 4.71 (2 decimal places)
2016-04-26 4:50 am
you have sin ( x / 2 ) = 1 / √2 ---> x/2 = π / 4 + 2nπ and 3π / 4 + 2 nπ...you finish
收錄日期: 2021-04-18 14:45:29
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