三角函数的题.麻烦了.Findalltheanglesbetween0°and360°:(1)sin(2x-30°)=cos30°(2)3sinYcosY=2cos^2Y(3)2tan^2Z+11secZ+7=0详细过程.麻烦了
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问题描述：
三角函数的题.麻烦了.
Findalltheanglesbetween0°and360°:
(1)sin(2x-30°)=cos30°
(2)3sinYcosY=2cos^2Y
(3)2tan^2Z+11secZ+7=0
详细过程.麻烦了
潘志庚回答：
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^2Y-3sinYcosY=0
cos^2Y(2-3tanY)=0
=>cosY=0ortanY=2/3Y∈[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=-1/2orsecZ=-5
cosZ=-2(del),-1/5Z∈[0,2π]
=>Z=arccos(-1/5),2π-arccos(-1/5)
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