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受欢迎的三角函数 >

csc(θ)-cot^2(θ)+1=0

解答

csc (θ) - cot^{2} (θ) + 1 = 0

解答

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

csc (θ) - cot^{2} (θ) + 1 = 0

使用三角恒等式改写

1 - cot^{2} (θ) + csc (θ)

使用毕达哥拉斯恒等式:

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

cot^{2} (x) = csc^{2} (x) - 1

= 1 - (csc^{2} (θ) - 1) + csc (θ)

化简

1 - (csc^{2} (θ) - 1) + csc (θ) : csc (θ) - csc^{2} (θ) + 2

1 - (csc^{2} (θ) - 1) + csc (θ)

- (csc^{2} (θ) - 1) : - csc^{2} (θ) + 1

- (csc^{2} (θ) - 1)

打开括号

= - (csc^{2} (θ)) - (- 1)

使用加减运算法则

- (- a) = a, - (a) = - a

= - csc^{2} (θ) + 1

= 1 - csc^{2} (θ) + 1 + csc (θ)

化简

1 - csc^{2} (θ) + 1 + csc (θ) : csc (θ) - csc^{2} (θ) + 2

1 - csc^{2} (θ) + 1 + csc (θ)

对同类项分组

= - csc^{2} (θ) + csc (θ) + 1 + 1

数字相加:

1 + 1 = 2

= csc (θ) - csc^{2} (θ) + 2

= csc (θ) - csc^{2} (θ) + 2

= csc (θ) - csc^{2} (θ) + 2

2 + csc (θ) - csc^{2} (θ) = 0

用替代法求解

2 + csc (θ) - csc^{2} (θ) = 0

令

: csc (θ) = u

2 + u - u^{2} = 0

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

2 + u - u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = 1, c = 2

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

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

1^{2} - 4 (- 1) \cdot 2 = 3

1^{2} - 4 (- 1) \cdot 2

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 ( - 1 )}, u_{2} = \frac{- 1 - 3}{2 ( - 1 )}

u = \frac{- 1 + 3}{2 ( - 1 )} : - 1

\frac{- 1 + 3}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 1}

数字相加/相减:

- 1 + 3 = 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

u = \frac{- 1 - 3}{2 ( - 1 )} : 2

\frac{- 1 - 3}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 1}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

二次方程组的解是:

u = - 1, u = 2

u = csc (θ)

代回

csc (θ) = - 1, csc (θ) = 2

csc (θ) = - 1, csc (θ) = 2

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 1

csc (θ) = - 1

的通解

csc (x)

周期表（周期为

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

csc (θ) = 2 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

csc (θ) = 2

csc (θ) = 2

的通解

csc (x)

周期表（周期为

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

合并所有解

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

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