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受欢迎的三角函数 >

cot^2(x)+2cot(x)+1=0

解答

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

解答

x = \frac{3 π}{4} + πn

+1

度数

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

求解步骤

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

用替代法求解

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

令

: cot (x) = u

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

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

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 \cdot 1 \cdot 1}{2 \cdot 1}

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

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

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 \cdot 1}

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

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

\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

(cot(x)+1)(sqrt(3)cot(x)+1)=0

(cot (x) + 1) (3 cot (x) + 1) = 0

sqrt(3)sin(θ)=-cos(θ)

3 sin (θ) = - cos (θ)

- 2 sin (x) + 1 = 0

sin(2θ)=-(sqrt(3))/2 ,0<= θ<= 2pi

sin (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

2-2cos^2(x/2)=2sin^2(x)

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