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受欢迎的 >

(sin(x)tan(x))/(1+cos(x))=tan(x)

解答

\frac{sin ( x ) tan ( x )}{1 + cos ( x )} = tan (x)

解答

x \in R 无解

求解步骤

\frac{sin ( x ) tan ( x )}{1 + cos ( x )} = tan (x)

两边减去

tan (x)

\frac{sin ( x ) tan ( x )}{1 + cos ( x )} - tan (x) = 0

化简

\frac{sin ( x ) tan ( x )}{1 + cos ( x )} - tan (x) : \frac{sin ( x ) tan ( x ) - tan ( x ) ( 1 + cos ( x ) )}{1 + cos ( x )}

\frac{sin ( x ) tan ( x )}{1 + cos ( x )} - tan (x)

将项转换为分式:

tan (x) = \frac{tan ( x ) ( 1 + cos ( x ) )}{1 + cos ( x )}

= \frac{sin ( x ) tan ( x )}{1 + cos ( x )} - \frac{tan ( x ) ( 1 + cos ( x ) )}{1 + cos ( 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 ( x ) tan ( x ) - tan ( x ) ( 1 + cos ( x ) )}{1 + cos ( x )}

\frac{sin ( x ) tan ( x ) - tan ( x ) ( 1 + cos ( x ) )}{1 + cos ( x )} = 0

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

sin (x) tan (x) - tan (x) (1 + cos (x)) = 0

分解

sin (x) tan (x) - tan (x) (1 + cos (x)) : tan (x) (sin (x) - 1 - cos (x))

sin (x) tan (x) - tan (x) (1 + cos (x))

因式分解出通项

tan (x)

= tan (x) (sin (x) - (1 + cos (x)))

乘开

sin (x) - (cos (x) + 1) : sin (x) - 1 - cos (x)

sin (x) - (1 + cos (x))

- (1 + cos (x)) : - 1 - cos (x)

- (1 + cos (x))

打开括号

= - (1) - (cos (x))

使用加减运算法则

+ (- a) = - a

= - 1 - cos (x)

= sin (x) - 1 - cos (x)

= tan (x) (sin (x) - cos (x) - 1)

tan (x) (sin (x) - 1 - cos (x)) = 0

分别求解每个部分

tan (x) = 0 or sin (x) - 1 - cos (x) = 0

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

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}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

sin (x) - 1 - cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = 2 πn + π

sin (x) - 1 - cos (x) = 0

使用三角恒等式改写

sin (x) - 1 - cos (x)

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

sin (x) - cos (x)

改写为

= 2 (\frac{1}{2} sin (x) - \frac{1}{2} cos (x))

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

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

使用角和恒等式:

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

= 2 sin (x - \frac{π}{4})

= - 1 + 2 sin (x - \frac{π}{4})

- 1 + 2 sin (x - \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (x - \frac{π}{4}) = 0

两边加上

1

- 1 + 2 sin (x - \frac{π}{4}) + 1 = 0 + 1

化简

2 sin (x - \frac{π}{4}) = 1

2 sin (x - \frac{π}{4}) = 1

两边除以

2

2 sin (x - \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{1}{2}

化简

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

\frac{2 sin ( x - \frac{π}{4} )}{2}

约分:

2

= sin (x - \frac{π}{4})

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

sin (x - \frac{π}{4}) = \frac{2}{2}

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}

x - \frac{π}{4} = \frac{π}{4} + 2 πn, x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

x - \frac{π}{4} = \frac{π}{4} + 2 πn, x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

解

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

x - \frac{π}{4} = \frac{π}{4} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{π}{4} + 2 πn

两边加上

\frac{π}{4}

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

化简

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

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{π}{4} + 2 πn + \frac{π}{4} : \frac{π}{2} + 2 πn

\frac{π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= \frac{π}{4} + \frac{π}{4} + 2 πn

合并分式

\frac{π}{4} + \frac{π}{4} : \frac{π}{2}

使用法则

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

= \frac{π + π}{4}

同类项相加:

π + π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

解

x - \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = 2 πn + π

x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{3 π}{4} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{3 π}{4} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{3 π}{4} + 2 πn + \frac{π}{4} : 2 πn + π

\frac{3 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{3 π}{4}

合并分式

\frac{π}{4} + \frac{3 π}{4} : π

使用法则

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

= \frac{π + 3 π}{4}

同类项相加:

π + 3 π = 4 π

= \frac{4 π}{4}

数字相除:

\frac{4}{4} = 1

= π

= 2 πn + π

x = 2 πn + π

x = 2 πn + π

x = 2 πn + π

x = \frac{π}{2} + 2 πn, x = 2 πn + π

合并所有解

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

因为方程对以下值无定义:

πn, \frac{π}{2} + 2 πn, 2 πn + π

x \in R 无解

流行的例子

sec (x + 4 5^{\circ}) = 1

117.72sin(θ)-35.316cos(θ)-12.5=0

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

- 2 sin (θ) = 0

sec^{3} (x) = - 1

1/(cot^2(x))+sqrt(3)tan(x)=0

\frac{1}{cot ^{2} ( x )} + 3 tan (x) = 0