题目

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

提问时间：2021-03-27

答案

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)