tan(2x-4 π)≤根号3

x=15+90k(k为整数),所以在0到360内有4个解:15,105,195,285.你的75,225是错的.再问：哪来的15+90K????再答：3tan2x-√3=0→tan2x=√3/32x=

tanx+tan(3/2π-x)=tanx+ctgx=sinx/cosx+cosx/sinx=1/sinx*cosx原式=sinx+sin(x-π/2)=根号2(sinx-cosx)平方=21-2*s

y=根号下tan(x/2+π/3)定义域为[kπ-π/3,kπ+2π/3)周期为π,在[kπ-π/3,kπ+2π/3)内单调递增

f(x)=1/cos²x-tan²x+根号2*sin(2x-π/4)=1/cos²x-sin²x/cos²x+根号2*sin(2x-π/4)=(1-s

首先我们知道tan(π/3)=sqrt(3)所以我们把原等式中的的sqrt(3)用tan(π/3)代替,即原式左边=[tan(π/3)-tan(π/3-x)]/[1+tan(π/3)tan(π/3-x

由tan(A+B)=(tanA+tanB)/(1-tanA*tanB)得,tan(18-x)tan(12+x)+tan(18-x)tan(12+x)+[tan(18-x)+tan(12+x)]=tan

1.1+tanx>=0tanx≥-1tan-45=-1在一个周期[-π/2,π/2]内的解集为x∈[-π/4,π/2]tanx又是一个周期为π的函数,所以x∈[nπ-π/4,nπ+π/2]（n∈z)2

已知tanx=根号2,x属于(π,3π/2),第三象限.sinx/cosx=根号2sinx=cosx根号2因为：sin²x+cos²x=1所以,3cos²x=1cosx=

a=(2cosX/2,tan(X/2+π/4))=（2cosx/2,（1+tanx/2）/（1-tanx/2））b=(根号二sin(X/2+π/4)),tan=(X/2—π/4)=（sinx/2+co

（1-根号3）/2

/>（1）tan(α+β-π/4)=tan(π/12+α+β-π/3)=[tan(π/12+α)+tan(β-π/3)]/[1-tan(π/12+α)tan(β-π/3)]=(√2+2√2)/[1-√

tan(π/6)=√3/3tan周期是π所以tan(kπ+π/6)=√3/3tan在一个周期(kπ-π/2,kπ+π/2)内是增函数tan(2x+π/3)>tan(kπ+π/6)所以kπ+π/6

f(x)=a●b=2cos(x/2)*√2*sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)=2√2sin(x/2+π/4)*cos(x/2)+tan(x/2+π/4)*t

f'(x)=(sinθ)x^2+((根号3)cosθ)x+tanθf'(π/4)=(sinθ)(π^2)/16+[(根号3)*(π/4)*cosθ]+tanθ题目很阴险啊,想让别人把θ和x弄混.

tan(2x-π/6)≥根号3kπ+π/2>2x-π/6≥π/3+kπkπ+2π/3>2x≥π/2+kπkπ/2+π/3>x≥π/4+kπ/2解集为{x|kπ/2+π/3>x≥π/4+kπ/2}

-tanθ≥√3tan-θ≥√3∴kπ+π/3≤-θ＜kπ+π/2∴-kπ-π/2＜θ≤-kπ-π/3

π/4-x=-kπ+arctan(√3/2)x=kπ+π/4-arctan(√3/2)

画y=tanx图像y-tanx在(kπ-π/2,kπ+π/2)上是增函数tan(x+π/6)≤根号3=tanπ/3kπ-π/2

tan(x+π/3)≥-根号3kπ-π/3≤x+π/3

tan(3x+π/4)=根号3,3x+π/4=kπ+π/6x=kπ/3-π/36,k∈Zx属于[0,2π],k=1,2,3,4,5,6符合条件的角的集合:{x|x=kπ/3-π/36,k=1,2,3,