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受欢迎的三角函数 >

csc^2(θ)+5=7-csc(θ)

解答

csc^{2} (θ) + 5 = 7 - csc (θ)

解答

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

csc^{2} (θ) + 5 = 7 - csc (θ)

用替代法求解

csc^{2} (θ) + 5 = 7 - csc (θ)

令

: csc (θ) = u

u^{2} + 5 = 7 - u

u^{2} + 5 = 7 - u : u = 1, u = - 2

u^{2} + 5 = 7 - u

将

u

para o lado esquerdo

u^{2} + 5 = 7 - u

两边加上

u

u^{2} + 5 + u = 7 - u + u

化简

u^{2} + 5 + u = 7

u^{2} + 5 + u = 7

将

7

para o lado esquerdo

u^{2} + 5 + u = 7

两边减去

7

u^{2} + 5 + u - 7 = 7 - 7

化简

u^{2} + u - 2 = 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 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 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 \cdot 1}

将解分隔开

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

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

\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}

使用法则

\frac{a}{a} = 1

= 1

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

\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{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

csc (θ) = - 2 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

流行的例子

3tan^2(x/3)-1=0

3 tan^{2} (\frac{x}{3}) - 1 = 0

cos (x) = \frac{7}{10}

2 cos (2 x) + 3 = 2

2 cos (x) = sec (x)

solvefor x,7y^2+sin(3x)=12-y^4

so l v e f or x, 7 y^{2} + sin (3 x) = 12 - y^{4}