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受欢迎的三角函数 >

4csc^2(θ)-25=0

解答

4 csc^{2} (θ) - 25 = 0

解答

θ = 0.41151 \dots + 2 πn, θ = π - 0.41151 \dots + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

+1

度数

θ = 23.57817 \dots^{\circ} + 36 0^{\circ} n, θ = 156.42182 \dots^{\circ} + 36 0^{\circ} n, θ = - 23.57817 \dots^{\circ} + 36 0^{\circ} n, θ = 203.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 csc^{2} (θ) - 25 = 0

用替代法求解

4 csc^{2} (θ) - 25 = 0

令

: csc (θ) = u

4 u^{2} - 25 = 0

4 u^{2} - 25 = 0 : u = \frac{5}{2}, u = - \frac{5}{2}

4 u^{2} - 25 = 0

将

25

到右边

4 u^{2} - 25 = 0

两边加上

25

4 u^{2} - 25 + 25 = 0 + 25

化简

4 u^{2} = 25

4 u^{2} = 25

两边除以

4

4 u^{2} = 25

两边除以

4

\frac{4 u ^{2}}{4} = \frac{25}{4}

化简

u^{2} = \frac{25}{4}

u^{2} = \frac{25}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{25}{4}, u = - \frac{25}{4}

\frac{25}{4} = \frac{5}{2}

\frac{25}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{25}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{25}{2}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{2}

- \frac{25}{4} = - \frac{5}{2}

- \frac{25}{4}

化简

\frac{25}{4} : \frac{5}{2}

\frac{25}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{25}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{25}{2}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{2}

= - \frac{5}{2}

u = \frac{5}{2}, u = - \frac{5}{2}

u = csc (θ)

代回

csc (θ) = \frac{5}{2}, csc (θ) = - \frac{5}{2}

csc (θ) = \frac{5}{2}, csc (θ) = - \frac{5}{2}

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

csc (θ) = \frac{5}{2}

使用反三角函数性质

csc (θ) = \frac{5}{2}

csc (θ) = \frac{5}{2}

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

θ = arccsc (\frac{5}{2}) + 2 πn, θ = π - arccsc (\frac{5}{2}) + 2 πn

θ = arccsc (\frac{5}{2}) + 2 πn, θ = π - arccsc (\frac{5}{2}) + 2 πn

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

csc (θ) = - \frac{5}{2}

使用反三角函数性质

csc (θ) = - \frac{5}{2}

csc (θ) = - \frac{5}{2}

的通解

csc (x) = - a \Rightarrow x = arccsc (- a) + 2 πn, x = π + arccsc (a) + 2 πn

θ = arccsc (- \frac{5}{2}) + 2 πn, θ = π + arccsc (\frac{5}{2}) + 2 πn

θ = arccsc (- \frac{5}{2}) + 2 πn, θ = π + arccsc (\frac{5}{2}) + 2 πn

合并所有解

θ = arccsc (\frac{5}{2}) + 2 πn, θ = π - arccsc (\frac{5}{2}) + 2 πn, θ = arccsc (- \frac{5}{2}) + 2 πn, θ = π + arccsc (\frac{5}{2}) + 2 πn

以小数形式表示解

θ = 0.41151 \dots + 2 πn, θ = π - 0.41151 \dots + 2 πn, θ = - 0.41151 \dots + 2 πn, θ = π + 0.41151 \dots + 2 πn

作图

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