三角函数的题.麻烦了.Find all the angles between 0°and 360°:(1)sin(2x-30°)=cos30°(2)3sinYcosY=2cos^2 Y(3)2tan^2 Z +11secZ+7=0详细过程.麻烦了
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![三角函数的题.麻烦了.Find all the angles between 0°and 360°:(1)sin(2x-30°)=cos30°(2)3sinYcosY=2cos^2 Y(3)2tan^2 Z +11secZ+7=0详细过程.麻烦了](/uploads/image/z/7131329-17-9.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E9%A2%98.%E9%BA%BB%E7%83%A6%E4%BA%86.Find+all+the+angles+between+0%C2%B0and+360%C2%B0%3A%281%29sin%282x-30%C2%B0%29%3Dcos30%C2%B0%282%293sinYcosY%3D2cos%5E2+Y%283%292tan%5E2+Z+%2B11secZ%2B7%3D0%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B.%E9%BA%BB%E7%83%A6%E4%BA%86)
三角函数的题.麻烦了.Find all the angles between 0°and 360°:(1)sin(2x-30°)=cos30°(2)3sinYcosY=2cos^2 Y(3)2tan^2 Z +11secZ+7=0详细过程.麻烦了
三角函数的题.麻烦了.
Find all the angles between 0°and 360°:
(1)sin(2x-30°)=cos30°
(2)3sinYcosY=2cos^2 Y
(3)2tan^2 Z +11secZ+7=0
详细过程.麻烦了
三角函数的题.麻烦了.Find all the angles between 0°and 360°:(1)sin(2x-30°)=cos30°(2)3sinYcosY=2cos^2 Y(3)2tan^2 Z +11secZ+7=0详细过程.麻烦了
sin(2x-30°)=sqrt(3)/2
x∈[0,2π] => (2x-30°)∈[-π/6,4π-π/6]
=> 2x-30°=mπ/3 + nπ,(m=1,2;n=1,2,3)
=> x=...
(2)
2cos^2 Y - 3sinYcosY = 0
cos^2Y(2-3tanY)=0
=> cosY=0 or tanY=2/3 Y∈[0,2π]
=> Y=π/2,arctan(2/3),arctan(2/3)+π
(3)
2(sec^2Z - 1) +11secZ+7=0
2sec^2Z +11secZ+5=0
(2secZ+1)(secZ+5)=0
secZ=-1/2 or secZ=-5
cosZ=-2(del),-1/5 Z∈[0,2π]
=> Z=arccos(-1/5),2π-arccos(-1/5)