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受欢迎的三角函数 >

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

解答

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

解答

x = 0, x = 2 π

+1

度数

x = 0^{\circ}, x = 36 0^{\circ}

求解步骤

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

两边减去

1

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

用 sin, cos 表示

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

化简

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

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

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 )} - 1

将项转换为分式:

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

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

乘以:

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

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

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

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

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

两边加上

cos (x)

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

两边进行平方

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

两边减去

cos^{2} (x)

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

使用三角恒等式改写

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

化简

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

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

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

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

1 - 1 = 0

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

- 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

使用指数法则:

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

若

n

是偶数

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

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} : 1

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

使用法则

- (- a) = a

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

使用法则

- (- a) = a

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

数字相减:

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

二次方程组的解是:

u = 1, u = 0

u = sin (x)

代回

sin (x) = 1, sin (x) = 0

sin (x) = 1, sin (x) = 0

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

sin (x) = 1, 0 \leq x \leq 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 \leq 2 π

范围内的解

x = \frac{π}{2}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = 0, x = π, x = 2 π

合并所有解

x = \frac{π}{2}, x = 0, x = π, x = 2 π

将解代入原方程进行验证

将它们代入

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

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) = 1

代入

x = \frac{π}{2}

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

未定义

\Rightarrow 假

检验

0

的解:

真

0

代入

n = 1

0

对于

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

代入

x = 0

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

整理后得

1 = 1

\Rightarrow 真

检验

π

的解:

假

π

代入

n = 1

π

对于

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

代入

x = π

- tan (π) + sec (π) = 1

整理后得

- 1 = 1

\Rightarrow 假

检验

2 π

的解:

真

2 π

代入

n = 1

2 π

对于

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

代入

x = 2 π

- tan (2 π) + sec (2 π) = 1

整理后得

1 = 1

\Rightarrow 真

x = 0, x = 2 π

作图

流行的例子

3tan(x/9)-sqrt(3)=0

solvefor x,tan(x)= 1/(sqrt(3))

2tan^2(x)+1=sec^2(x)+sin^2(x)+cos^2(x)