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受欢迎的三角函数 >

tan(x)-sec(x)=3,0<= x<2pi

解答

tan (x) - sec (x) = 3, 0 \leq x < 2 π

解答

x = π + 0.92729 \dots

+1

度数

x = 233.13010 \dots^{\circ}

求解步骤

tan (x) - sec (x) = 3, 0 \leq x < 2 π

两边减去

3

tan (x) - sec (x) - 3 = 0

用 sin, cos 表示

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

化简

\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} - 3 : \frac{sin ( x ) - 1 - 3 cos ( x )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} - 3

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 )}

= \frac{sin ( x ) - 1}{cos ( x )} - 3

将项转换为分式:

3 = \frac{3 cos ( x )}{cos ( x )}

= \frac{sin ( x ) - 1}{cos ( x )} - \frac{3 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{sin ( x ) - 1 - 3 cos ( x )}{cos ( x )}

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

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

sin (x) - 1 - 3 cos (x) = 0

两边加上

3 cos (x)

sin (x) - 1 = 3 cos (x)

两边进行平方

(sin (x) - 1)^{2} = (3 cos (x))^{2}

两边减去

(3 cos (x))^{2}

(sin (x) - 1)^{2} - 9 cos^{2} (x) = 0

使用三角恒等式改写

(- 1 + sin (x))^{2} - 9 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= (- 1 + sin (x))^{2} - 9 (1 - sin^{2} (x))

化简

(- 1 + sin (x))^{2} - 9 (1 - sin^{2} (x)) : 10 sin^{2} (x) - 2 sin (x) - 8

(- 1 + sin (x))^{2} - 9 (1 - sin^{2} (x))

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

使用完全平方公式:

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

a = - 1, b = sin (x)

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

化简

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

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

去除括号:

(- a) = - a

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

2 \cdot 1 \cdot sin (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin (x)

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

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

= 1 - 2 sin (x) + sin^{2} (x) - 9 (1 - sin^{2} (x))

乘开

- 9 (1 - sin^{2} (x)) : - 9 + 9 sin^{2} (x)

- 9 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (x)

= - 9 \cdot 1 - (- 9) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (x)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (x)

= 1 - 2 sin (x) + sin^{2} (x) - 9 + 9 sin^{2} (x)

化简

1 - 2 sin (x) + sin^{2} (x) - 9 + 9 sin^{2} (x) : 10 sin^{2} (x) - 2 sin (x) - 8

1 - 2 sin (x) + sin^{2} (x) - 9 + 9 sin^{2} (x)

对同类项分组

= - 2 sin (x) + sin^{2} (x) + 9 sin^{2} (x) + 1 - 9

同类项相加:

sin^{2} (x) + 9 sin^{2} (x) = 10 sin^{2} (x)

= - 2 sin (x) + 10 sin^{2} (x) + 1 - 9

数字相加/相减:

1 - 9 = - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

= 10 sin^{2} (x) - 2 sin (x) - 8

- 8 + 10 sin^{2} (x) - 2 sin (x) = 0

用替代法求解

- 8 + 10 sin^{2} (x) - 2 sin (x) = 0

令

: sin (x) = u

- 8 + 10 u^{2} - 2 u = 0

- 8 + 10 u^{2} - 2 u = 0 : u = 1, u = - \frac{4}{5}

- 8 + 10 u^{2} - 2 u = 0

改写成标准形式

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

10 u^{2} - 2 u - 8 = 0

使用求根公式求解

10 u^{2} - 2 u - 8 = 0

二次方程求根公式:

若

a = 10, b = - 2, c = - 8

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

(- 2)^{2} - 4 \cdot 10 (- 8) = 18

(- 2)^{2} - 4 \cdot 10 (- 8)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 10 \cdot 8

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 10 \cdot 8

数字相乘:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

数字相加:

4 + 320 = 324

= 324

因式分解数字:

324 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 2 ) \pm 18}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 18}{2 \cdot 10}, u_{2} = \frac{- ( - 2 ) - 18}{2 \cdot 10}

u = \frac{- ( - 2 ) + 18}{2 \cdot 10} : 1

\frac{- ( - 2 ) + 18}{2 \cdot 10}

使用法则

- (- a) = a

= \frac{2 + 18}{2 \cdot 10}

数字相加:

2 + 18 = 20

= \frac{20}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{20}{20}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 18}{2 \cdot 10} : - \frac{4}{5}

\frac{- ( - 2 ) - 18}{2 \cdot 10}

使用法则

- (- a) = a

= \frac{2 - 18}{2 \cdot 10}

数字相减:

2 - 18 = - 16

= \frac{- 16}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 16}{20}

使用分式法则:

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

= - \frac{16}{20}

约分:

4

= - \frac{4}{5}

二次方程组的解是:

u = 1, u = - \frac{4}{5}

u = sin (x)

代回

sin (x) = 1, sin (x) = - \frac{4}{5}

sin (x) = 1, sin (x) = - \frac{4}{5}

sin (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x < 2 π

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}

sin (x) = - \frac{4}{5}, 0 \leq x < 2 π : x = π + arcsin (\frac{4}{5}), x = - arcsin (\frac{4}{5}) + 2 π

sin (x) = - \frac{4}{5}, 0 \leq x < 2 π

使用反三角函数性质

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

在

0 \leq x < 2 π

范围内的解

x = π + arcsin (\frac{4}{5}), x = - arcsin (\frac{4}{5}) + 2 π

合并所有解

x = \frac{π}{2}, x = π + arcsin (\frac{4}{5}), x = - arcsin (\frac{4}{5}) + 2 π

将解代入原方程进行验证

将它们代入

tan (x) - sec (x) = 3

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

tan (x) - sec (x) = 3

代入

x = \frac{π}{2}

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

未定义

\Rightarrow 假

检验

π + arcsin (\frac{4}{5})

的解:

真

π + arcsin (\frac{4}{5})

代入

n = 1

π + arcsin (\frac{4}{5})

对于

tan (x) - sec (x) = 3

代入

x = π + arcsin (\frac{4}{5})

tan (π + arcsin (\frac{4}{5})) - sec (π + arcsin (\frac{4}{5})) = 3

整理后得

3 = 3

\Rightarrow 真

检验

- arcsin (\frac{4}{5}) + 2 π

的解:

假

- arcsin (\frac{4}{5}) + 2 π

代入

n = 1

- arcsin (\frac{4}{5}) + 2 π

对于

tan (x) - sec (x) = 3

代入

x = - arcsin (\frac{4}{5}) + 2 π

tan (- arcsin (\frac{4}{5}) + 2 π) - sec (- arcsin (\frac{4}{5}) + 2 π) = 3

整理后得

- 3 = 3

\Rightarrow 假

x = π + arcsin (\frac{4}{5})

以小数形式表示解

x = π + 0.92729 \dots

作图

流行的例子

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3 (2 + cos (x)) = 5

tan(x)= 5/(8.66)

tan (x) = \frac{5}{8.66}

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