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受欢迎的三角函数 >

csc(x)=-sqrt(1-cot(x))

解答

csc (x) = - 1 - cot (x)

解答

x = \frac{π}{2} + πn, x = \frac{3 π}{4} + πn

+1

度数

x = 9 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

csc (x) = - 1 - cot (x)

两边进行平方

csc^{2} (x) = (- 1 - cot (x))^{2}

两边减去

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

csc^{2} (x) - 1 + cot (x) = 0

使用三角恒等式改写

- 1 + cot (x) + csc^{2} (x)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

csc^{2} (x) - 1 = cot^{2} (x)

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

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

用替代法求解

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

令

: cot (x) = u

u + u^{2} = 0

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 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} : 0

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

数字相加/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

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

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

数字相减:

- 1 - 1 = - 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 = 0, u = - 1

u = cot (x)

代回

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

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

cot (x) = 0 : x = \frac{π}{2} + πn

cot (x) = 0

cot (x) = 0

的通解

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

cot (x) = - 1

的通解

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = \frac{π}{2} + πn, x = \frac{3 π}{4} + πn

将解代入原方程进行验证

将它们代入

csc (x) = - 1 - cot (x)

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + πn

的解:

真

\frac{π}{2} + πn

代入

n = 1

\frac{π}{2} + π 1

对于

csc (x) = - 1 - cot (x)

代入

x = \frac{π}{2} + π 1

csc (\frac{π}{2} + π 1) = - 1 - cot (\frac{π}{2} + π 1)

整理后得

- 1 = - 1

\Rightarrow 真

检验

\frac{3 π}{4} + πn

的解:

真

\frac{3 π}{4} + πn

代入

n = 1

\frac{3 π}{4} + π 1

对于

csc (x) = - 1 - cot (x)

代入

x = \frac{3 π}{4} + π 1

csc (\frac{3 π}{4} + π 1) = - 1 - cot (\frac{3 π}{4} + π 1)

整理后得

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow 真

x = \frac{π}{2} + πn, x = \frac{3 π}{4} + πn

作图

流行的例子

cos(2θ)+5cos(θ)+3=0

cos (2 θ) + 5 cos (θ) + 3 = 0

2 cos (x) = 2

sin(2θ)=sqrt(3)sin(θ)

sin (2 θ) = 3 sin (θ)

cot (\frac{x}{2}) = 1

-8tan^2(θ)-4tan(θ)-11=6tan(θ)-8

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8