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受欢迎的三角函数 >

sqrt(3)cot(2x)+3=0

解答

3 cot (2 x) + 3 = 0

解答

x = \frac{5 π}{12} + \frac{πn}{2}

+1

度数

x = 7 5^{\circ} + 9 0^{\circ} n

求解步骤

3 cot (2 x) + 3 = 0

将

3

到右边

3 cot (2 x) + 3 = 0

两边减去

3

3 cot (2 x) + 3 - 3 = 0 - 3

化简

3 cot (2 x) = - 3

3 cot (2 x) = - 3

两边除以

3

3 cot (2 x) = - 3

两边除以

3

\frac{3 cot ( 2 x )}{3} = \frac{- 3}{3}

化简

\frac{3 cot ( 2 x )}{3} = \frac{- 3}{3}

化简

\frac{3 cot ( 2 x )}{3} : cot (2 x)

\frac{3 cot ( 2 x )}{3}

约分:

3

= cot (2 x)

化简

\frac{- 3}{3} : - 3

\frac{- 3}{3}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{3}

使用根式运算法则:

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

3 = 3^{\frac{1}{2}}

= \frac{3}{3 ^{\frac{1}{2}}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= 3^{1 - \frac{1}{2}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= 3^{\frac{1}{2}}

使用根式运算法则:

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

3^{\frac{1}{2}} = 3

= - 3

cot (2 x) = - 3

cot (2 x) = - 3

cot (2 x) = - 3

cot (2 x) = - 3

的通解

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

2 x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{6} + πn

解

2 x = \frac{5 π}{6} + πn : x = \frac{5 π}{12} + \frac{πn}{2}

2 x = \frac{5 π}{6} + πn

两边除以

2

2 x = \frac{5 π}{6} + πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

作图

流行的例子

\frac{1}{cos ( θ )} = 2

cos(θ)= 7/25 ,(3pi)/2 <θ<2pi

cos (θ) = \frac{7}{25}, \frac{3 π}{2} < θ < 2 π

solvefor x,cos(x)=-1

so l v e f or x, cos (x) = - 1

5sin(x)+1=3sin(x)

5 sin (x) + 1 = 3 sin (x)

3sin(x)+4=5cos(x)

3 sin (x) + 4 = 5 cos (x)