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受欢迎的三角函数 >

-2cot(x)=csc^2(x)

解答

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

解答

x = \frac{3 π}{4} + πn

+1

度数

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

求解步骤

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

两边减去

csc^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数字相减:

4 - 4 = 0

= 0

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

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

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

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

去除括号:

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

= \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

u = - 1

二次方程组的解是:

u = - 1

u = cot (x)

代回

cot (x) = - 1

cot (x) = - 1

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{3 π}{4} + πn

作图

流行的例子

sin (θ) = - 0.671

(sin^2(x)+5cos^2(x))/2 =2

\frac{sin ^{2} ( x ) + 5 cos ^{2} ( x )}{2} = 2

2sin(x)=7-3csc(x)

2 sin (x) = 7 - 3 csc (x)

cos(θ)=(134/109)

cos (θ) = (\frac{134}{109})

cos (2 x + 3 0^{\circ}) = 0