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受欢迎的三角函数 >

sec(x)-tan(x)=2

解答

sec (x) - tan (x) = 2

解答

x = - 0.64350 \dots + 2 πn

+1

度数

x = - 36.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sec (x) - tan (x) = 2

两边减去

2

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

用 sin, cos 表示

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

化简

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

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

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

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

两边加上

2 cos (x)

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

两边进行平方

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

两边减去

(2 cos (x))^{2}

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

使用三角恒等式改写

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

化简

(1 - sin (x))^{2} - 4 (1 - sin^{2} (x)) : 5 sin^{2} (x) - 2 sin (x) - 3

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

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

使用加减运算法则

- (- a) = a

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

1 - 2 sin (x) + sin^{2} (x) - 4 + 4 sin^{2} (x) : 5 sin^{2} (x) - 2 sin (x) - 3

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

对同类项分组

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

同类项相加:

sin^{2} (x) + 4 sin^{2} (x) = 5 sin^{2} (x)

= - 2 sin (x) + 5 sin^{2} (x) + 1 - 4

数字相加/相减:

1 - 4 = - 3

= 5 sin^{2} (x) - 2 sin (x) - 3

= 5 sin^{2} (x) - 2 sin (x) - 3

= 5 sin^{2} (x) - 2 sin (x) - 3

- 3 - 2 sin (x) + 5 sin^{2} (x) = 0

用替代法求解

- 3 - 2 sin (x) + 5 sin^{2} (x) = 0

令

: sin (x) = u

- 3 - 2 u + 5 u^{2} = 0

- 3 - 2 u + 5 u^{2} = 0 : u = 1, u = - \frac{3}{5}

- 3 - 2 u + 5 u^{2} = 0

改写成标准形式

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

5 u^{2} - 2 u - 3 = 0

使用求根公式求解

5 u^{2} - 2 u - 3 = 0

二次方程求根公式:

若

a = 5, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 2)^{2} - 4 \cdot 5 (- 3) = 8

(- 2)^{2} - 4 \cdot 5 (- 3)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 5 \cdot 3

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 5 \cdot 3

数字相乘:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

数字相加:

4 + 60 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 5}, u_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 5}

u = \frac{- ( - 2 ) + 8}{2 \cdot 5} : 1

\frac{- ( - 2 ) + 8}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 + 8}{2 \cdot 5}

数字相加:

2 + 8 = 10

= \frac{10}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10}{10}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 8}{2 \cdot 5} : - \frac{3}{5}

\frac{- ( - 2 ) - 8}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 - 8}{2 \cdot 5}

数字相减:

2 - 8 = - 6

= \frac{- 6}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 6}{10}

使用分式法则:

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

= - \frac{6}{10}

约分:

2

= - \frac{3}{5}

二次方程组的解是:

u = 1, u = - \frac{3}{5}

u = sin (x)

代回

sin (x) = 1, sin (x) = - \frac{3}{5}

sin (x) = 1, sin (x) = - \frac{3}{5}

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) = - \frac{3}{5} : x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

sin (x) = - \frac{3}{5}

使用反三角函数性质

sin (x) = - \frac{3}{5}

sin (x) = - \frac{3}{5}

的通解

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

x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = arcsin (- \frac{3}{5}) + 2 πn, x = π + arcsin (\frac{3}{5}) + 2 πn

将解代入原方程进行验证

将它们代入

sec (x) - tan (x) = 2

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) - tan (x) = 2

代入

x = \frac{π}{2} + 2 π 1

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

未定义

\Rightarrow 假

检验

arcsin (- \frac{3}{5}) + 2 πn

的解:

真

arcsin (- \frac{3}{5}) + 2 πn

代入

n = 1

arcsin (- \frac{3}{5}) + 2 π 1

对于

sec (x) - tan (x) = 2

代入

x = arcsin (- \frac{3}{5}) + 2 π 1

sec (arcsin (- \frac{3}{5}) + 2 π 1) - tan (arcsin (- \frac{3}{5}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

π + arcsin (\frac{3}{5}) + 2 πn

的解:

假

π + arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

π + arcsin (\frac{3}{5}) + 2 π 1

对于

sec (x) - tan (x) = 2

代入

x = π + arcsin (\frac{3}{5}) + 2 π 1

sec (π + arcsin (\frac{3}{5}) + 2 π 1) - tan (π + arcsin (\frac{3}{5}) + 2 π 1) = 2

整理后得

- 2 = 2

\Rightarrow 假

x = arcsin (- \frac{3}{5}) + 2 πn

以小数形式表示解

x = - 0.64350 \dots + 2 πn

作图

流行的例子

sin((2pi(t+101.75))/(365))=-1

sin (\frac{2 π ( t + 101.75 )}{365}) = - 1

cos(θ)=-(sqrt(7))/5

cos (θ) = - \frac{7}{5}

sin(pi+x)=sin(x)

sin (π + x) = sin (x)

cos (x) = - \frac{2}{π}

5 sin (x) = - 3