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受欢迎的三角函数 >

1+2cot(θ)-2cot^2(θ)=-3cot^2(θ)

解答

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

解答

θ = \frac{3 π}{4} + πn

+1

度数

θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cot (θ) = u

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

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

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

将

3 u^{2}

para o lado esquerdo

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

两边加上

3 u^{2}

1 + 2 u - 2 u^{2} + 3 u^{2} = - 3 u^{2} + 3 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 (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(x)-9=cos(x)-9

sin (x) - 9 = cos (x) - 9

7sin(x)+1=5sin(x)

7 sin (x) + 1 = 5 sin (x)

sec(x)+tan(x)=1,(0,2pi)

sec (x) + tan (x) = 1, (0, 2 π)

cos^2(θ)-cos(θ)-20=0

cos^{2} (θ) - cos (θ) - 20 = 0

100.0191=180-arctan((3.9254)/(0.6936-x))

100.0191 = 180 - arctan (\frac{3.9254}{0.6936 - x})