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受欢迎的三角函数 >

cot^2(θ)=-cot(θ)

解答

cot^{2} (θ) = - cot (θ)

解答

θ = \frac{π}{2} + πn, θ = \frac{3 π}{4} + πn

+1

度数

θ = 9 0^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

cot^{2} (θ) = - cot (θ)

用替代法求解

cot^{2} (θ) = - cot (θ)

令

: cot (θ) = u

u^{2} = - u

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

u^{2} = - u

将

u

para o lado esquerdo

u^{2} = - u

两边加上

u

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

化简

u^{2} + u = 0

u^{2} + u = 0

使用求根公式求解

u^{2} + u = 0

二次方程求根公式:

若

a = 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

u_{1} = \frac{- 1 + 1}{2 \cdot 1}, u_{2} = \frac{- 1 - 1}{2 \cdot 1}

u = \frac{- 1 + 1}{2 \cdot 1} : 0

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

数字相减:

- 1 - 1 = - 2

= \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

二次方程组的解是:

u = 0, u = - 1

u = cot (θ)

代回

cot (θ) = 0, cot (θ) = - 1

cot (θ) = 0, cot (θ) = - 1

cot (θ) = 0 : θ = \frac{π}{2} + πn

cot (θ) = 0

cot (θ) = 0

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{π}{2} + πn

θ = \frac{π}{2} + πn

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{π}{2} + πn, θ = \frac{3 π}{4} + πn

作图

流行的例子

0.2727=sec(2.746x)-1

0.2727 = sec (2.746 x) - 1

2sqrt(3)sin(2θ)=sqrt(3),0<= θ<2pi

23 sin (2 θ) = 3, 0 \leq θ < 2 π

cos(x)-sqrt(3)=3cos(x)

cos (x) - 3 = 3 cos (x)

2tan^2(x)-7sec(x)+5=0

2 tan^{2} (x) - 7 sec (x) + 5 = 0

cos(w)=0.75sin(90-w)

cos (w) = 0.75 sin (9 0^{\circ} - w)