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-1-tan(x)=sec(x),0<= x<2pi

解答

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

解答

x = π

+1

度数

x = 18 0^{\circ}

求解步骤

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

两边减去

sec (x)

- 1 - tan (x) - sec (x) = 0

用 sin, cos 表示

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

化简

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

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

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

两边加上

sin (x)

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

两边进行平方

(- cos (x) - 1)^{2} = sin^{2} (x)

两边减去

sin^{2} (x)

(- cos (x) - 1)^{2} - sin^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= (- 1 - cos (x))^{2} - (1 - cos^{2} (x))

化简

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

(- 1 - cos (x))^{2} - (1 - cos^{2} (x))

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

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = - 1, b = cos (x)

= (- 1)^{2} - 2 (- 1) cos (x) + cos^{2} (x)

化简

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

(- 1)^{2} - 2 (- 1) cos (x) + cos^{2} (x)

使用法则

- (- a) = a

= (- 1)^{2} + 2 \cdot 1 \cdot cos (x) + cos^{2} (x)

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 \cdot cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

= 1 + 2 cos (x) + cos^{2} (x)

= 1 + 2 cos (x) + cos^{2} (x)

= 1 + 2 cos (x) + cos^{2} (x) - (1 - cos^{2} (x))

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

- (1 - cos^{2} (x))

打开括号

= - (1) - (- cos^{2} (x))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (x)

= 1 + 2 cos (x) + cos^{2} (x) - 1 + cos^{2} (x)

化简

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

1 + 2 cos (x) + cos^{2} (x) - 1 + cos^{2} (x)

对同类项分组

= 2 cos (x) + cos^{2} (x) + cos^{2} (x) + 1 - 1

同类项相加:

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

= 2 cos (x) + 2 cos^{2} (x) + 1 - 1

1 - 1 = 0

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

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

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

2 cos (x) + 2 cos^{2} (x) = 0

用替代法求解

2 cos (x) + 2 cos^{2} (x) = 0

令

: cos (x) = u

2 u + 2 u^{2} = 0

2 u + 2 u^{2} = 0 : u = 0, u = - 1

2 u + 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} + 2 u = 0

使用求根公式求解

2 u^{2} + 2 u = 0

二次方程求根公式:

若

a = 2, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

2^{2} - 4 \cdot 2 \cdot 0 = 2

2^{2} - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 2 + 2}{2 \cdot 2}, u_{2} = \frac{- 2 - 2}{2 \cdot 2}

u = \frac{- 2 + 2}{2 \cdot 2} : 0

\frac{- 2 + 2}{2 \cdot 2}

数字相加/相减:

- 2 + 2 = 0

= \frac{0}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 2} : - 1

\frac{- 2 - 2}{2 \cdot 2}

数字相减:

- 2 - 2 = - 4

= \frac{- 4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 0, u = - 1

u = cos (x)

代回

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

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

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

cos (x) = - 1

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

在

0 \leq x < 2 π

范围内的解

x = π

合并所有解

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

将解代入原方程进行验证

将它们代入

- 1 - tan (x) = sec (x)

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2}

的解:

假

\frac{π}{2}

代入

n = 1

\frac{π}{2}

对于

- 1 - tan (x) = sec (x)

代入

x = \frac{π}{2}

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

整理后得

- \infty = \infty

\Rightarrow 假

检验

\frac{3 π}{2}

的解:

真

\frac{3 π}{2}

代入

n = 1

\frac{3 π}{2}

对于

- 1 - tan (x) = sec (x)

代入

x = \frac{3 π}{2}

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

整理后得

- \infty = - \infty

\Rightarrow 真

检验

π

的解:

真

π

代入

n = 1

π

对于

- 1 - tan (x) = sec (x)

代入

x = π

- 1 - tan (π) = sec (π)

整理后得

- 1 = - 1

\Rightarrow 真

x = \frac{3 π}{2}, x = π

因为方程对以下值无定义:

\frac{3 π}{2}

x = π

作图

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tan (x) = 11, sin (x)

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