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受欢迎的三角函数 >

-2cot(θ)=cot^2(θ)+1

解答

- 2 cot (θ) = cot^{2} (θ) + 1

解答

θ = \frac{3 π}{4} + πn

+1

度数

θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

- 2 cot (θ) = cot^{2} (θ) + 1

用替代法求解

- 2 cot (θ) = cot^{2} (θ) + 1

令

: cot (θ) = u

- 2 u = u^{2} + 1

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

- 2 u = u^{2} + 1

交换两边

u^{2} + 1 = - 2 u

将

2 u

para o lado esquerdo

u^{2} + 1 = - 2 u

两边加上

2 u

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

化简

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

u^{2} + 1 + 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 \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 (θ)

代回

cot (θ) = - 1

cot (θ) = - 1

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

cot (θ) = - 1

cot (θ) = - 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

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

合并所有解

θ = \frac{3 π}{4} + πn

作图

流行的例子

[sin(4x)cos(x)-cos(4x)sin(x)]^2=1

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

cos(θ)=(202)/(sqrt(331)*2\sqrt{33)}

cos (θ) = \frac{202}{331 \cdot 2 33}

5sin(θ)+12cos(θ)=13

5 sin (θ) + 12 cos (θ) = 13

2-2cos^2(x)+3cos(x)=0

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

cot(x)= 5/12 ,pi<x<(3pi)/2

cot (x) = \frac{5}{12}, π < x < \frac{3 π}{2}