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受欢迎的三角函数 >

-4cot^2(x)=1+2cot(x)-3cot^2(x)

解答

- 4 cot^{2} (x) = 1 + 2 cot (x) - 3 cot^{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

- 4 cot^{2} (x) = 1 + 2 cot (x) - 3 cot^{2} (x)

用替代法求解

- 4 cot^{2} (x) = 1 + 2 cot (x) - 3 cot^{2} (x)

令

: cot (x) = u

- 4 u^{2} = 1 + 2 u - 3 u^{2}

- 4 u^{2} = 1 + 2 u - 3 u^{2} : u = - 1

- 4 u^{2} = 1 + 2 u - 3 u^{2}

交换两边

1 + 2 u - 3 u^{2} = - 4 u^{2}

将

4 u^{2}

para o lado esquerdo

1 + 2 u - 3 u^{2} = - 4 u^{2}

两边加上

4 u^{2}

1 + 2 u - 3 u^{2} + 4 u^{2} = - 4 u^{2} + 4 u^{2}

化简

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

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

tan(x+pi/4)-tan(x-pi/4)=3

tan (x + \frac{π}{4}) - tan (x - \frac{π}{4}) = 3

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

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

2sin^2(θ)-3sin(θ)=-1

2 sin^{2} (θ) - 3 sin (θ) = - 1

cos (kπ) = 0

sin (θ) = - 0.8