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受欢迎的三角函数 >

3tan^2(2x)=1

解答

3 tan^{2} (2 x) = 1

解答

x = \frac{π}{12} + \frac{πn}{2}, x = - \frac{π}{12} + \frac{πn}{2}

+1

度数

x = 1 5^{\circ} + 9 0^{\circ} n, x = - 1 5^{\circ} + 9 0^{\circ} n

求解步骤

3 tan^{2} (2 x) = 1

用替代法求解

3 tan^{2} (2 x) = 1

令

: tan (2 x) = u

3 u^{2} = 1

3 u^{2} = 1 : u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{2} = 1

两边除以

3

3 u^{2} = 1

两边除以

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

u = tan (2 x)

代回

tan (2 x) = \frac{1}{3}, tan (2 x) = - \frac{1}{3}

tan (2 x) = \frac{1}{3}, tan (2 x) = - \frac{1}{3}

tan (2 x) = \frac{1}{3} : x = \frac{π}{12} + \frac{πn}{2}

tan (2 x) = \frac{1}{3}

使用反三角函数性质

tan (2 x) = \frac{1}{3}

tan (2 x) = \frac{1}{3}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

2 x = arctan (\frac{1}{3}) + πn

2 x = arctan (\frac{1}{3}) + πn

解

2 x = arctan (\frac{1}{3}) + πn : x = \frac{π}{12} + \frac{πn}{2}

2 x = arctan (\frac{1}{3}) + πn

化简

arctan (\frac{1}{3}) + πn : \frac{π}{6} + πn

arctan (\frac{1}{3}) + πn

使用以下普通恒等式:

arctan (\frac{1}{3}) = \frac{π}{6}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{6} + πn

2 x = \frac{π}{6} + πn

两边除以

2

2 x = \frac{π}{6} + πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{πn}{2} : \frac{π}{12} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

tan (2 x) = - \frac{1}{3} : x = - \frac{π}{12} + \frac{πn}{2}

tan (2 x) = - \frac{1}{3}

使用反三角函数性质

tan (2 x) = - \frac{1}{3}

tan (2 x) = - \frac{1}{3}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

2 x = arctan (- \frac{1}{3}) + πn

2 x = arctan (- \frac{1}{3}) + πn

解

2 x = arctan (- \frac{1}{3}) + πn : x = - \frac{π}{12} + \frac{πn}{2}

2 x = arctan (- \frac{1}{3}) + πn

化简

arctan (- \frac{1}{3}) + πn : - \frac{π}{6} + πn

arctan (- \frac{1}{3}) + πn

arctan (- \frac{1}{3}) = - \frac{π}{6}

arctan (- \frac{1}{3})

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- \frac{1}{3}) = - arctan (\frac{1}{3})

= - arctan (\frac{1}{3})

使用以下普通恒等式:

arctan (\frac{1}{3}) = \frac{π}{6}

arctan (\frac{1}{3})

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= - \frac{π}{6} + πn

2 x = - \frac{π}{6} + πn

两边除以

2

2 x = - \frac{π}{6} + πn

两边除以

2

\frac{2 x}{2} = - \frac{\frac{π}{6}}{2} + \frac{πn}{2}

化简

\frac{2 x}{2} = - \frac{\frac{π}{6}}{2} + \frac{πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- \frac{\frac{π}{6}}{2} + \frac{πn}{2} : - \frac{π}{12} + \frac{πn}{2}

- \frac{\frac{π}{6}}{2} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

= - \frac{π}{12} + \frac{πn}{2}

x = - \frac{π}{12} + \frac{πn}{2}

x = - \frac{π}{12} + \frac{πn}{2}

x = - \frac{π}{12} + \frac{πn}{2}

x = - \frac{π}{12} + \frac{πn}{2}

合并所有解

x = \frac{π}{12} + \frac{πn}{2}, x = - \frac{π}{12} + \frac{πn}{2}

作图

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