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受欢迎的三角函数 >

2sec(x)+2tan(x)=2

解答

2 sec (x) + 2 tan (x) = 2

解答

x = 2 πn + 2 π

+1

度数

x = 36 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sec (x) + 2 tan (x) = 2

两边减去

2

2 sec (x) + 2 tan (x) - 2 = 0

用 sin, cos 表示

2 \cdot \frac{1}{cos ( x )} + 2 \cdot \frac{sin ( x )}{cos ( x )} - 2 = 0

化简

2 \cdot \frac{1}{cos ( x )} + 2 \cdot \frac{sin ( x )}{cos ( x )} - 2 : \frac{2 + 2 sin ( x ) - 2 cos ( x )}{cos ( x )}

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

2 \cdot \frac{1}{cos ( x )} = \frac{2}{cos ( x )}

2 \cdot \frac{1}{cos ( x )}

分式相乘:

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

= \frac{1 \cdot 2}{cos ( x )}

数字相乘:

1 \cdot 2 = 2

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

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

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

分式相乘:

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

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

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

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

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

两边加上

2 cos (x)

2 + 2 sin (x) = 2 cos (x)

两边进行平方

(2 + 2 sin (x))^{2} = (2 cos (x))^{2}

两边减去

(2 cos (x))^{2}

(2 + 2 sin (x))^{2} - 4 cos^{2} (x) = 0

分解

(2 + 2 sin (x))^{2} - 4 cos^{2} (x) : 4 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x))

(2 + 2 sin (x))^{2} - 4 cos^{2} (x)

将

(2 + 2 sin (x))^{2} - 4 cos^{2} (x)

改写为

(2 + 2 sin (x))^{2} - (2 cos (x))^{2}

(2 + 2 sin (x))^{2} - 4 cos^{2} (x)

将

4

改写为

2^{2}

= (2 + 2 sin (x))^{2} - 2^{2} cos^{2} (x)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

2^{2} cos^{2} (x) = (2 cos (x))^{2}

= (2 + 2 sin (x))^{2} - (2 cos (x))^{2}

= (2 + 2 sin (x))^{2} - (2 cos (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 + 2 sin (x))^{2} - (2 cos (x))^{2} = ((2 + 2 sin (x)) + 2 cos (x)) ((2 + 2 sin (x)) - 2 cos (x))

= ((2 + 2 sin (x)) + 2 cos (x)) ((2 + 2 sin (x)) - 2 cos (x))

整理后得

= (2 sin (x) + 2 cos (x) + 2) (2 sin (x) - 2 cos (x) + 2)

分解

2 + 2 sin (x) + 2 cos (x) : 2 (1 + sin (x) + cos (x))

2 + 2 sin (x) + 2 cos (x)

因式分解出通项

2

= 2 (1 + sin (x) + cos (x))

= 2 (sin (x) + cos (x) + 1) (2 sin (x) - 2 cos (x) + 2)

分解

2 + 2 sin (x) - 2 cos (x) : 2 (1 + sin (x) - cos (x))

2 + 2 sin (x) - 2 cos (x)

因式分解出通项

2

= 2 (1 + sin (x) - cos (x))

= 2 (1 + sin (x) + cos (x)) \cdot 2 (1 + sin (x) - cos (x))

整理后得

= 4 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x))

4 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x)) = 0

分别求解每个部分

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

1 + sin (x) + cos (x) = 0 : x = 2 πn + π, x = 2 πn + \frac{3 π}{2}

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

使用三角恒等式改写

1 + sin (x) + 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

两边减去

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}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \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}

= - \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{5 π}{4} + 2 πn, x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

x + \frac{π}{4} = \frac{5 π}{4} + 2 πn, x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

解

x + \frac{π}{4} = \frac{5 π}{4} + 2 πn : x = 2 πn + π

x + \frac{π}{4} = \frac{5 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = \frac{5 π}{4} + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{4} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} = \frac{5 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{5 π}{4} + 2 πn - \frac{π}{4} : 2 πn + π

\frac{5 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{5 π}{4}

合并分式

- \frac{π}{4} + \frac{5 π}{4} : π

使用法则

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

= \frac{- π + 5 π}{4}

同类项相加:

- π + 5 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

解

x + \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = 2 πn + \frac{3 π}{2}

x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = \frac{7 π}{4} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} = \frac{7 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{7 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{7 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{7 π}{4}

合并分式

- \frac{π}{4} + \frac{7 π}{4} : \frac{3 π}{2}

使用法则

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

= \frac{- π + 7 π}{4}

同类项相加:

- π + 7 π = 6 π

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

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

1 + sin (x) - cos (x) = 0 : x = 2 πn + \frac{3 π}{2}, x = 2 πn + 2 π

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

使用三角恒等式改写

1 + sin (x) - 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

两边减去

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}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \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}

= - \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{5 π}{4} + 2 πn, x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn, x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

解

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn : x = 2 πn + \frac{3 π}{2}

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{4} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{5 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{5 π}{4} + 2 πn + \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{5 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{5 π}{4}

合并分式

\frac{π}{4} + \frac{5 π}{4} : \frac{3 π}{2}

使用法则

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

= \frac{π + 5 π}{4}

同类项相加:

π + 5 π = 6 π

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

解

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = 2 πn + 2 π

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{7 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{7 π}{4} + 2 πn + \frac{π}{4} : 2 πn + 2 π

\frac{7 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{7 π}{4}

合并分式

\frac{π}{4} + \frac{7 π}{4} : 2 π

使用法则

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

= \frac{π + 7 π}{4}

同类项相加:

π + 7 π = 8 π

= \frac{8 π}{4}

数字相除:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

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

合并所有解

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

将解代入原方程进行验证

将它们代入

2 sec (x) + 2 tan (x) = 2

检验解是否符合

去除与方程不符的解。

检验

2 πn + π

的解:

假

2 πn + π

代入

n = 1

2 π 1 + π

对于

2 sec (x) + 2 tan (x) = 2

代入

x = 2 π 1 + π

2 sec (2 π 1 + π) + 2 tan (2 π 1 + π) = 2

整理后得

- 2 = 2

\Rightarrow 假

检验

2 πn + \frac{3 π}{2}

的解:

假

2 πn + \frac{3 π}{2}

代入

n = 1

2 π 1 + \frac{3 π}{2}

对于

2 sec (x) + 2 tan (x) = 2

代入

x = 2 π 1 + \frac{3 π}{2}

2 sec (2 π 1 + \frac{3 π}{2}) + 2 tan (2 π 1 + \frac{3 π}{2}) = 2

未定义

\Rightarrow 假

检验

2 πn + 2 π

的解:

真

2 πn + 2 π

代入

n = 1

2 π 1 + 2 π

对于

2 sec (x) + 2 tan (x) = 2

代入

x = 2 π 1 + 2 π

2 sec (2 π 1 + 2 π) + 2 tan (2 π 1 + 2 π) = 2

整理后得

2 = 2

\Rightarrow 真

x = 2 πn + 2 π

作图

流行的例子

2sin^2(θ)+sqrt(3)sin(θ)=0

2 sin^{2} (θ) + 3 sin (θ) = 0

cos^2(x)=sin^2(x/2)

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

sin(x+pi/4)=-(sqrt(2))/2

sin (x + \frac{π}{4}) = - \frac{2}{2}

1 = tan (θ)

2 cos (x) - 4 = 0