Question

求函数定义域，值域

f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))

Answer 1

1+2sin(2x+π/3)≠0，解得2x+π/3≠2kπ+7π/6 且 2x+π/3≠2kπ+11π/6
x≠kπ-π/4 且 x≠kπ+5π/12
f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))
设2x+π/3=θ+π/6
f(x)=(√3-2cos(θ+π/6))/(1+2sin(θ+π/6))
=(√3-√3cosθ+sinθ)/(1+√3sinθ+cosθ)
∵sinθ/(1+cosθ)=(1-cosθ)/sinθ=tan(θ/2)
∴由和比定理得f(x)=tan(θ/2)=tan(x+π/12)
∵x≠kπ-π/4 且 x≠kπ+5π/12
∴f(x)≠-√3/3

Answer 2

解：要使函数有意义，则
1+2sin(2x+π/3)≠0
sin(2x+π/3)≠1/2
2x+π/3≠2kπ+π/6且2x+π/3≠2kπ+5π/6
解得;x≠kπ-π/12 且 x≠kπ+π/4 (k属于Z)
√3-2cos(2x+π/3)=√3-2[cos²(x+π/6)-sin²(x+π/6)]
=√3[(cos²(x+π/6)+sin²(x+π/6)]-2[cos²(x+π/6)-sin²(x+π/6)]
=(√3-2)cos²(x+π/6)+(√3+2)sin²(x+π/6)
1+2sin(2x+π/3)=1+4sin(x+π/6)cos(x+π/6)
=cos²(x+π/6)+sin²(x+π/6)+4sin(x+π/6)cos(x+π/6)
y=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))
=[(√3-2)cos²(x+π/6)+(√3+2)sin²(x+π/6)]/[cos²(x+π/6)+sin²(x+π/6)+4sin(x+π/6)cos(x+π/6)]
=[√3-2+(√3+2)tan²(x+π/6)]/[tan²(x+π/6)+4tan(x+π/6)+1]
所以，[tan²(x+π/6)+4tan(x+π/6)+1]y=[√3-2+(√3+2)tan²(x+π/6)]
移项：(2+√3-y)tan²(x+π/6)-4ytan(x+π/6)-y-2+√3=0
Δ=b²-4ac>=0
即：(-4y)²-4(2+√3-y)(-y-2+√3)>=0
解得：12y²+8√3y+4>=0
4(√3y+1)²>=0
等式恒成立，故y属于R

Answer 3

1+2sin(2x+π/3)≠0, sin(2x+π/3)≠-1/2
2x+π/3≠2kπ-π/6且2x+π/3≠2kπ-5π/6
解得;x≠kπ-π/4 且 x≠kπ-7π/12 (k属于Z)
y=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))=(√3/2-cost)/(1/2+sint)表示圆上两点A(-sint,coct)和B(1/2,√3/2)连线的斜率（A，B不重合），A，B重合时表示切线的斜率，故y≠-√3/3

Answer 4

解：函数解析式含有分式，自变量只需满足分母不等于0就可以了
即1+2sin(2x+π/3)<>0，
sin(2x+π/3)<>-1/2
2x+π/3<>2kπ+7π/6 且 2x+π/3<>2kπ-π/6
解得x<>kπ+5π/12且x<>kπ-π/4

追问

值域？还有，能化简不？