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受欢迎的三角函数 >

25cot^2(x)-1=0

解答

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

解答

x = 1.37340 \dots + πn, x = 1.76819 \dots + πn

+1

度数

x = 78.69006 \dots^{\circ} + 18 0^{\circ} n, x = 101.30993 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cot (x) = u

25 u^{2} - 1 = 0

25 u^{2} - 1 = 0 : u = \frac{1}{5}, u = - \frac{1}{5}

25 u^{2} - 1 = 0

将

1

到右边

25 u^{2} - 1 = 0

两边加上

1

25 u^{2} - 1 + 1 = 0 + 1

化简

25 u^{2} = 1

25 u^{2} = 1

两边除以

25

25 u^{2} = 1

两边除以

25

\frac{25 u ^{2}}{25} = \frac{1}{25}

化简

u^{2} = \frac{1}{25}

u^{2} = \frac{1}{25}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{25}, u = - \frac{1}{25}

\frac{1}{25} = \frac{1}{5}

\frac{1}{25}

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{1}{25}

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{1}{5}

使用法则

1 = 1

= \frac{1}{5}

- \frac{1}{25} = - \frac{1}{5}

- \frac{1}{25}

化简

\frac{1}{25} : \frac{1}{5}

\frac{1}{25}

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{1}{25}

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{1}{5}

= - \frac{1}{5}

使用法则

1 = 1

= - \frac{1}{5}

u = \frac{1}{5}, u = - \frac{1}{5}

u = cot (x)

代回

cot (x) = \frac{1}{5}, cot (x) = - \frac{1}{5}

cot (x) = \frac{1}{5}, cot (x) = - \frac{1}{5}

cot (x) = \frac{1}{5} : x = arccot (\frac{1}{5}) + πn

cot (x) = \frac{1}{5}

使用反三角函数性质

cot (x) = \frac{1}{5}

cot (x) = \frac{1}{5}

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1}{5}) + πn

x = arccot (\frac{1}{5}) + πn

cot (x) = - \frac{1}{5} : x = arccot (- \frac{1}{5}) + πn

cot (x) = - \frac{1}{5}

使用反三角函数性质

cot (x) = - \frac{1}{5}

cot (x) = - \frac{1}{5}

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- \frac{1}{5}) + πn

x = arccot (- \frac{1}{5}) + πn

合并所有解

x = arccot (\frac{1}{5}) + πn, x = arccot (- \frac{1}{5}) + πn

以小数形式表示解

x = 1.37340 \dots + πn, x = 1.76819 \dots + πn

作图

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