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受欢迎的三角函数 >

sec^2(x)cot(x)=4

解答

sec^{2} (x) cot (x) = 4

解答

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

+1

度数

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) cot (x) = 4

两边减去

4

sec^{2} (x) cot (x) - 4 = 0

用 sin, cos 表示

- 4 + cot (x) sec^{2} (x)

使用基本三角恒等式:

cot (x) = \frac{cos ( x )}{sin ( x )}

= - 4 + \frac{cos ( x )}{sin ( x )} sec^{2} (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - 4 + \frac{cos ( x )}{sin ( x )} (\frac{1}{cos ( x )})^{2}

化简

- 4 + \frac{cos ( x )}{sin ( x )} (\frac{1}{cos ( x )})^{2} : \frac{- 4 cos ( x ) sin ( x ) + 1}{cos ( x ) sin ( x )}

- 4 + \frac{cos ( x )}{sin ( x )} (\frac{1}{cos ( x )})^{2}

\frac{cos ( x )}{sin ( x )} (\frac{1}{cos ( x )})^{2} = \frac{1}{cos ( x ) sin ( x )}

\frac{cos ( x )}{sin ( x )} (\frac{1}{cos ( x )})^{2}

分式相乘:

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

= \frac{cos ( x ) ( \frac{1}{c o s ( x )} ) ^{2}}{sin ( x )}

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

= \frac{\frac{1}{c o s ^{2} ( x )} cos ( x )}{sin ( x )}

乘

cos (x) \frac{1}{cos ^{2} ( x )} : \frac{1}{cos ( x )}

cos (x) \frac{1}{cos ^{2} ( x )}

分式相乘:

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

= \frac{1 \cdot cos ( x )}{cos ^{2} ( x )}

乘以:

1 \cdot cos (x) = cos (x)

= \frac{cos ( x )}{cos ^{2} ( x )}

约分:

cos (x)

= \frac{1}{cos ( x )}

= \frac{\frac{1}{c o s ( x )}}{sin ( x )}

使用分式法则:

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

= \frac{1}{cos ( x ) sin ( x )}

= - 4 + \frac{1}{cos ( x ) sin ( x )}

将项转换为分式:

4 = \frac{4 cos ( x ) sin ( x )}{cos ( x ) sin ( x )}

= - \frac{4 cos ( x ) sin ( x )}{cos ( x ) sin ( x )} + \frac{1}{cos ( x ) sin ( x )}

因为分母相等，所以合并分式:

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

= \frac{- 4 cos ( x ) sin ( x ) + 1}{cos ( x ) sin ( x )}

= \frac{- 4 cos ( x ) sin ( x ) + 1}{cos ( x ) sin ( x )}

\frac{1 - 4 cos ( x ) sin ( x )}{cos ( x ) sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - 4 cos (x) sin (x) = 0

使用三角恒等式改写

1 - 4 cos (x) sin (x)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= 1 - 4 \cdot \frac{sin ( 2 x )}{2}

1 - 4 \cdot \frac{sin ( 2 x )}{2} = 0

4 \cdot \frac{sin ( 2 x )}{2} = 2 sin (2 x)

4 \cdot \frac{sin ( 2 x )}{2}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 sin (2 x)

1 - 2 sin (2 x) = 0

将

1

到右边

1 - 2 sin (2 x) = 0

两边减去

1

1 - 2 sin (2 x) - 1 = 0 - 1

化简

- 2 sin (2 x) = - 1

- 2 sin (2 x) = - 1

两边除以

- 2

- 2 sin (2 x) = - 1

两边除以

- 2

\frac{- 2 sin ( 2 x )}{- 2} = \frac{- 1}{- 2}

化简

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

解

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

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

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

化简

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

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

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

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

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

化简

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

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

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

作图

流行的例子

1=sqrt(4cos(2θ))

1 = 4 cos (2 θ)

209cos(47)=150cos(θ)

209 cos (4 7^{\circ}) = 150 cos (θ)

7-5cos(x)=10,7-5cos(x)=10

7 - 5 cos (x) = 10, 7 - 5 cos (x) = 10

cos (4 x) = cos (x)

-4cos(2θ)-19=-26cos(θ),0<= θ<360

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}