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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

两边减去

cot^{2} (x)

1 + csc (x) - cot^{2} (x) = 0

使用三角恒等式改写

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

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} (x) - 1) + csc (x)

化简

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

1 - (csc^{2} (x) - 1) + csc (x)

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

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

打开括号

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

使用加减运算法则

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

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

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

化简

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

1 - csc^{2} (x) + 1 + csc (x)

对同类项分组

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

数字相加:

1 + 1 = 2

= csc (x) - csc^{2} (x) + 2

= csc (x) - csc^{2} (x) + 2

= csc (x) - csc^{2} (x) + 2

2 + csc (x) - csc^{2} (x) = 0

用替代法求解

2 + csc (x) - csc^{2} (x) = 0

令

: csc (x) = 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 (x)

代回

csc (x) = - 1, csc (x) = 2

csc (x) = - 1, csc (x) = 2

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

csc (x) = - 1

csc (x) = - 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

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

csc (x) = 2

csc (x) = 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

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

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

合并所有解

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

作图

流行的例子

cos(1/(4θ))=(-sqrt(2))/2

cos (\frac{1}{4 θ}) = \frac{- 2}{2}

cos(x)tan(x)-2cos^2(x)=-1

cos (x) tan (x) - 2 cos^{2} (x) = - 1

sqrt(3)cot(1/4 x)=1

3 cot (\frac{1}{4} x) = 1

2sin(pi/4-x)+sqrt(3)=0

2 sin (\frac{π}{4} - x) + 3 = 0

3.1=48.1(1-cos((28.66x)/(48.1)))

3.1 = 48.1 (1 - cos (\frac{28.66 x}{48.1}))