Question

解方程，求x的数值

Answer 1

tanx+tan3x=√3(1-tanxtan3x)
∴tan4x=tan(x+3x)
=(tanx+tan3x)/(1-tanxtan3x)
=√3
∴4x=π/3或4π/3或7π/3或10π/3
∴x=π/12或π/3或7π/12或5π/6

Answer 2

tanx+tan3x=√3[1-tan3xtanx]
(tanx+tan3x)/(1-tan3xtanx)=√3=tan(π/3+kπ)
tan4x=tan(π/3+kπ)
4x=π/3+kπ
x=π/12+kπ/4
0≤π/12+kπ/4<π
0≤(1/12)+(k/4)<1
0≤(1/3)+k<4
-(1/3)≤k<4-(1/3)
-(1/3)≤k<11/3
k=0, 1 , 2 , 3
x1=π/12
x2=π/3
x3=7π/12
x4=5π/6

Answer 3

tanx+tan3x=√3(1-tanxtan3x)
(tanx+tan3x)/(1-tanxtan3x)=√3
tan4x=√3

∵0≤x＜π
∴0≤4x＜4π
∴4x=π/3，或4π/3，或7π/3，或10π/3
∴x=π/12，或π/3，或7π/12，或5π/6

Answer 4