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受欢迎的三角函数 >

tan(4x)-tan(2x)=0

解答

tan (4 x) - tan (2 x) = 0

解答

x = πn, x = \frac{π}{2} + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

tan (4 x) - tan (2 x) = 0

用 sin, cos 表示

- tan (2 x) + tan (4 x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - \frac{sin ( 2 x )}{cos ( 2 x )} + tan (4 x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - \frac{sin ( 2 x )}{cos ( 2 x )} + \frac{sin ( 4 x )}{cos ( 4 x )}

化简

- \frac{sin ( 2 x )}{cos ( 2 x )} + \frac{sin ( 4 x )}{cos ( 4 x )} : \frac{- sin ( 2 x ) cos ( 4 x ) + sin ( 4 x ) cos ( 2 x )}{cos ( 2 x ) cos ( 4 x )}

- \frac{sin ( 2 x )}{cos ( 2 x )} + \frac{sin ( 4 x )}{cos ( 4 x )}

cos (2 x), cos (4 x)

的最小公倍数:

cos (2 x) cos (4 x)

cos (2 x), cos (4 x)

最小公倍数 (LCM)

计算出由出现在

cos (2 x)

或

cos (4 x)

中的因子组成的表达式

= cos (2 x) cos (4 x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (2 x) cos (4 x)

对于

\frac{sin ( 2 x )}{cos ( 2 x )} :

将分母和分子乘以

cos (4 x)

\frac{sin ( 2 x )}{cos ( 2 x )} = \frac{sin ( 2 x ) cos ( 4 x )}{cos ( 2 x ) cos ( 4 x )}

对于

\frac{sin ( 4 x )}{cos ( 4 x )} :

将分母和分子乘以

cos (2 x)

\frac{sin ( 4 x )}{cos ( 4 x )} = \frac{sin ( 4 x ) cos ( 2 x )}{cos ( 4 x ) cos ( 2 x )}

= - \frac{sin ( 2 x ) cos ( 4 x )}{cos ( 2 x ) cos ( 4 x )} + \frac{sin ( 4 x ) cos ( 2 x )}{cos ( 4 x ) cos ( 2 x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- sin ( 2 x ) cos ( 4 x ) + sin ( 4 x ) cos ( 2 x )}{cos ( 2 x ) cos ( 4 x )}

= \frac{- sin ( 2 x ) cos ( 4 x ) + sin ( 4 x ) cos ( 2 x )}{cos ( 2 x ) cos ( 4 x )}

\frac{cos ( 2 x ) sin ( 4 x ) - cos ( 4 x ) sin ( 2 x )}{cos ( 2 x ) cos ( 4 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (2 x) sin (4 x) - cos (4 x) sin (2 x) = 0

使用三角恒等式改写

cos (2 x) sin (4 x) - cos (4 x) sin (2 x)

使用角差恒等式:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (4 x - 2 x)

sin (4 x - 2 x) = 0

sin (4 x - 2 x) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

4 x - 2 x = 0 + 2 πn, 4 x - 2 x = π + 2 πn

4 x - 2 x = 0 + 2 πn, 4 x - 2 x = π + 2 πn

解

4 x - 2 x = 0 + 2 πn : x = πn

4 x - 2 x = 0 + 2 πn

同类项相加:

4 x - 2 x = 2 x

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

解

4 x - 2 x = π + 2 πn : x = \frac{π}{2} + πn

4 x - 2 x = π + 2 πn

同类项相加:

4 x - 2 x = 2 x

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

作图

流行的例子

solvefor t,sin(t)+2sin(2t)=0

so l v e f or t, sin (t) + 2 sin (2 t) = 0

tan^2(x)-sin(x)tan^2(x)=0

tan^{2} (x) - sin (x) tan^{2} (x) = 0

4sin(x)=2.54648

4 sin (x) = 2.54648

5sec^3(x)-20sec(x)=0

5 sec^{3} (x) - 20 sec (x) = 0

solvefor 3cot^2(x)-1=0,x

so l v e f or 3 cot^{2} (x) - 1 = 0, x