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受欢迎的三角函数 >

3sqrt(cot(x-10))=4

解答

3 cot (x - 1 0^{\circ}) = 4

解答

x = 0.51238 \dots + 18 0^{\circ} n + 1 0^{\circ}

+1

弧度

x = 0.51238 \dots + \frac{π}{18} + πn

求解步骤

3 cot (x - 1 0^{\circ}) = 4

两边除以

3

3 cot (x - 1 0^{\circ}) = 4

两边除以

3

\frac{3 cot ( x - 1 0 ^{\circ} )}{3} = \frac{4}{3}

化简

cot (x - 1 0^{\circ}) = \frac{4}{3}

cot (x - 1 0^{\circ}) = \frac{4}{3}

两边进行平方:

cot (x - 1 0^{\circ}) = \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{4}{3}

(cot (x - 1 0^{\circ}))^{2} = (\frac{4}{3})^{2}

展开

(cot (x - 1 0^{\circ}))^{2} : cot (x - 1 0^{\circ})

(cot (x - 1 0^{\circ}))^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (cot^{\frac{1}{2}} (x - 1 0^{\circ}))^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= cot^{\frac{1}{2} \cdot 2} (x - 1 0^{\circ})

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

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

2

= 1

= cot (x - 1 0^{\circ})

展开

(\frac{4}{3})^{2} : \frac{16}{9}

(\frac{4}{3})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{4 ^{2}}{3 ^{2}}

4^{2} = 16

= \frac{16}{3 ^{2}}

3^{2} = 9

= \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{16}{9}

验证解:

cot (x - 1 0^{\circ}) = \frac{16}{9}

真

将它们代入

3 cot (x - 1 0^{\circ}) = 4

检验解是否符合

去除与方程不符的解。

代入

cot (x - 1 0^{\circ}) = \frac{16}{9} :

真

3 \frac{16}{9} = 4

3 \frac{16}{9} = 4

3 \frac{16}{9}

\frac{16}{9} = \frac{4}{3}

\frac{16}{9}

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{16}{9}

9 = 3

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

= \frac{16}{3}

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

= \frac{4}{3}

= 3 \cdot \frac{4}{3}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 \cdot 3}{3}

约分:

3

= 4

4 = 4

真

解是

cot (x - 1 0^{\circ}) = \frac{16}{9}

使用反三角函数性质

cot (x - 1 0^{\circ}) = \frac{16}{9}

cot (x - 1 0^{\circ}) = \frac{16}{9}

的通解

cot (x) = a \Rightarrow x = arccot (a) + 18 0^{\circ} n

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

解

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n : x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

将

1 0^{\circ}

到右边

x - 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n

两边加上

1 0^{\circ}

x - 1 0^{\circ} + 1 0^{\circ} = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

化简

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

x = arccot (\frac{16}{9}) + 18 0^{\circ} n + 1 0^{\circ}

以小数形式表示解

x = 0.51238 \dots + 18 0^{\circ} n + 1 0^{\circ}

作图

流行的例子

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2.4638 cos (x) = 2.5843

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