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受欢迎的三角函数 >

sec(x)-(sin(x))/(cos(x))=cos(x)

解答

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

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cos (x)

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

化简

sec (x) - \frac{sin ( x )}{cos ( x )} - cos (x) : \frac{sec ( x ) cos ( x ) - sin ( x ) - cos ^{2} ( x )}{cos ( x )}

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

将项转换为分式:

sec (x) = \frac{sec ( x ) cos ( x )}{cos ( x )}, cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}

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

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

sec (x) cos (x) - sin (x) - cos (x) cos (x)

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

cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

用 sin, cos 表示

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

化简

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

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

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

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

分式相乘:

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

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

约分:

cos (x)

= 1

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

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

两边加上

cos^{2} (x)

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

两边进行平方

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

两边减去

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

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

分解

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

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

使用指数法则:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

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

使用平方差公式:

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

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

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

整理后得

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

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

分别求解每个部分

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

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

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

使用三角恒等式改写

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

化简

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

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

用替代法求解

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

令

: sin (x) = u

2 - u - u^{2} = 0

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

2 - u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = - 1, c = 2

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

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

(- 1)^{2} - 4 (- 1) \cdot 2

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 2

(- 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

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 1 )} : - 2

\frac{- ( - 1 ) + 3}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 3}{- 2 \cdot 1}

数字相加:

1 + 3 = 4

= \frac{4}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{- 2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

u = \frac{- ( - 1 ) - 3}{2 ( - 1 )} : 1

\frac{- ( - 1 ) - 3}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 3}{- 2 \cdot 1}

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - 2, u = 1

u = sin (x)

代回

sin (x) = - 2, sin (x) = 1

sin (x) = - 2, sin (x) = 1

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

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

合并所有解

x = \frac{π}{2} + 2 πn

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (x) = u

- u + u^{2} = 0

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

- u + u^{2} = 0

改写成标准形式

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

u^{2} - u = 0

使用求根公式求解

u^{2} - u = 0

二次方程求根公式:

若

a = 1, b = - 1, c = 0

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

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

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 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

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

\frac{- ( - 1 ) + 1}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

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 : x = \frac{π}{2} + 2 πn

sin (x) = 1

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

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

合并所有解

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

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

假

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

x = \frac{π}{2} + 2 π 1

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

未定义

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

x = 2 πn, x = π + 2 πn

作图

流行的例子

cos(θ)=0.4,sec(θ)

cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ))

sec(3x)-csc(30)=0,(x+35)/5

4cos(2θ)-10cos(θ)+14=7,0<= θ<360