三角函数的题。。帮忙解决一下。。谢谢

Find all the angles between 0°and 360°:(1)4sin^2 X=6-9cosX(2)3cosY+cotY=0
2025-03-31 09:51:26
推荐回答(1个)
回答1:

(1)4(1-cos^2 X)=6-9cosX
4cos^2 X-9cosX-2=0
cosX=[9-sqrt(113)]/8 记作A,另一解大于1,舍去。
x=arccosy,x∈[0,π]
=>X=arccosA,2π-arccosA

(2)3cosY+cosY/sinY=0
sinY不等于0,两边乘sinY
=>3cosYsinY+cosY=0
=>cosY(3sinY+1)=0
=>cosY=0或者sinY=-1/3 又y=arcsinx,y∈[-π/2,π/2]
所以,Y=π/2,arcsin(-1/3)+2π,π-arcsin(-1/3)