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受欢迎的 >

sqrt(3)tan(2x-pi)+1=0

解答

3 tan (2 x - π) + 1 = 0

解答

x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

+1

度数

x = 16 5^{\circ} + 9 0^{\circ} n

求解步骤

3 tan (2 x - π) + 1 = 0

将

1

到右边

3 tan (2 x - π) + 1 = 0

两边减去

1

3 tan (2 x - π) + 1 - 1 = 0 - 1

化简

3 tan (2 x - π) = - 1

3 tan (2 x - π) = - 1

两边除以

3

3 tan (2 x - π) = - 1

两边除以

3

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

化简

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

化简

\frac{3 tan ( 2 x - π )}{3} : tan (2 x - π)

\frac{3 tan ( 2 x - π )}{3}

约分:

3

= tan (2 x - π)

化简

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

- \frac{1}{3}

有理化:

- \frac{3}{3}

- \frac{1}{3}

乘以共轭根式

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (2 x - π) = - \frac{3}{3}

tan (2 x - π) = - \frac{3}{3}

tan (2 x - π) = - \frac{3}{3}

tan (2 x - π) = - \frac{3}{3}

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x - π = \frac{5 π}{6} + πn

2 x - π = \frac{5 π}{6} + πn

解

2 x - π = \frac{5 π}{6} + πn : x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

2 x - π = \frac{5 π}{6} + πn

将

π

到右边

2 x - π = \frac{5 π}{6} + πn

两边加上

π

2 x - π + π = \frac{5 π}{6} + πn + π

化简

2 x = \frac{5 π}{6} + πn + π

2 x = \frac{5 π}{6} + πn + π

两边除以

2

2 x = \frac{5 π}{6} + πn + π

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2} + \frac{π}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2} + \frac{π}{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{5 π}{6}}{2} + \frac{πn}{2} + \frac{π}{2} : \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2} + \frac{π}{2}

对同类项分组

= \frac{π}{2} + \frac{πn}{2} + \frac{\frac{5 π}{6}}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

= \frac{π}{2} + \frac{πn}{2} + \frac{5 π}{12}

对同类项分组

= \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{2} + \frac{5 π}{12} + \frac{πn}{2}

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solvefor x,2sin(x)+csc(x)=0 求解

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