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受欢迎的三角函数 >

2cot(x)+sec^2(x)=0

解答

2 cot (x) + sec^{2} (x) = 0

解答

x = \frac{3 π}{4} + πn

+1

度数

x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

2 cot (x) + sec^{2} (x) = 0

使用三角恒等式改写

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

使用基本三角恒等式:

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

= sec^{2} (x) + 2 \cdot \frac{1}{tan ( x )}

2 \cdot \frac{1}{tan ( x )} = \frac{2}{tan ( x )}

2 \cdot \frac{1}{tan ( x )}

分式相乘:

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

= \frac{1 \cdot 2}{tan ( x )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{tan ( x )}

= sec^{2} (x) + \frac{2}{tan ( x )}

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

= \frac{2}{tan ( x )} + tan^{2} (x) + 1

1 + \frac{2}{tan ( x )} + tan^{2} (x) = 0

用替代法求解

1 + \frac{2}{tan ( x )} + tan^{2} (x) = 0

令

: tan (x) = u

1 + \frac{2}{u} + u^{2} = 0

1 + \frac{2}{u} + u^{2} = 0 : u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

1 + \frac{2}{u} + u^{2} = 0

在两边乘以

u

1 + \frac{2}{u} + u^{2} = 0

在两边乘以

u

1 \cdot u + \frac{2}{u} u + u^{2} u = 0 \cdot u

化简

1 \cdot u + \frac{2}{u} u + u^{2} u = 0 \cdot u

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= u

化简

\frac{2}{u} u : 2

\frac{2}{u} u

分式相乘:

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

= \frac{2 u}{u}

约分:

u

= 2

化简

u^{2} u : u^{3}

u^{2} u

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= u^{2 + 1}

数字相加:

2 + 1 = 3

= u^{3}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

u + 2 + u^{3} = 0

u + 2 + u^{3} = 0

u + 2 + u^{3} = 0

解

u + 2 + u^{3} = 0 : u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u + 2 + u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{3} + u + 2 = 0

因式分解

u^{3} + u + 2 : (u + 1) (u^{2} - u + 2)

u^{3} + u + 2

使用有理根定理

a_{0} = 2, a_{n} = 1

a_{0}

的除数

: 1, 2, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1 , 2}{1}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{u ^{3} + u + 2}{u + 1}

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

\frac{u ^{3} + u + 2}{u + 1}

对

\frac{u ^{3} + u + 2}{u + 1}

做除法:

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

将分子

u^{3} + u + 2

与除数

u + 1

的首项系数相除：

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

商 = u^{2}

将

u + 1

乘以

u^{2} : u^{3} + u^{2}

将

u^{3} + u + 2

减去

u^{3} + u^{2}

得到新的余数

余数 = - u^{2} + u + 2

因此

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

= u^{2} + \frac{- u ^{2} + u + 2}{u + 1}

对

\frac{- u ^{2} + u + 2}{u + 1}

做除法:

\frac{- u ^{2} + u + 2}{u + 1} = - u + \frac{2 u + 2}{u + 1}

将分子

- u^{2} + u + 2

与除数

u + 1

的首项系数相除：

\frac{- u ^{2}}{u} = - u

商 = - u

将

u + 1

乘以

- u : - u^{2} - u

将

- u^{2} + u + 2

减去

- u^{2} - u

得到新的余数

余数 = 2 u + 2

因此

\frac{- u ^{2} + u + 2}{u + 1} = - u + \frac{2 u + 2}{u + 1}

= u^{2} - u + \frac{2 u + 2}{u + 1}

对

\frac{2 u + 2}{u + 1}

做除法:

\frac{2 u + 2}{u + 1} = 2

将分子

2 u + 2

与除数

u + 1

的首项系数相除：

\frac{2 u}{u} = 2

商 = 2

将

u + 1

乘以

2 : 2 u + 2

将

2 u + 2

减去

2 u + 2

得到新的余数

余数 = 0

因此

\frac{2 u + 2}{u + 1} = 2

= u^{2} - u + 2

= (u + 1) (u^{2} - u + 2)

(u + 1) (u^{2} - u + 2) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u^{2} - u + 2 = 0 : u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u^{2} - u + 2 = 0

使用求根公式求解

u^{2} - u + 2 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = 2

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

化简

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

(- 1)^{2} - 4 \cdot 1 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 - 8

数字相减:

1 - 8 = - 7

= - 7

使用根式运算法则:

- a = - 1 a

- 7 = - 1 7

= - 1 7

使用虚数运算法则:

- 1 = i

= 7 i

u_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

u = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

\frac{- ( - 1 ) + 7 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 7 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 7 i}{2}

将

\frac{1 + 7 i}{2}

改写成标准复数形式:

\frac{1}{2} + \frac{7}{2} i

\frac{1 + 7 i}{2}

使用分式法则:

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

\frac{1 + 7 i}{2} = \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7}{2} i

u = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

\frac{- ( - 1 ) - 7 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 7 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 7 i}{2}

将

\frac{1 - 7 i}{2}

改写成标准复数形式:

\frac{1}{2} - \frac{7}{2} i

\frac{1 - 7 i}{2}

使用分式法则:

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

\frac{1 - 7 i}{2} = \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7}{2} i

二次方程组的解是:

u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

解为

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

1 + \frac{2}{u} + u^{2}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = - 1, u = \frac{1}{2} + i \frac{7}{2}, u = \frac{1}{2} - i \frac{7}{2}

u = tan (x)

代回

tan (x) = - 1, tan (x) = \frac{1}{2} + i \frac{7}{2}, tan (x) = \frac{1}{2} - i \frac{7}{2}

tan (x) = - 1, tan (x) = \frac{1}{2} + i \frac{7}{2}, tan (x) = \frac{1}{2} - i \frac{7}{2}

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

tan (x) = - 1

的通解

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) = \frac{1}{2} + i \frac{7}{2} :

无解

tan (x) = \frac{1}{2} + i \frac{7}{2}

无解

tan (x) = \frac{1}{2} - i \frac{7}{2} :

无解

tan (x) = \frac{1}{2} - i \frac{7}{2}

无解

合并所有解

x = \frac{3 π}{4} + πn

作图

流行的例子

sin(x)+sin(x/2)=0

sin (x) + sin (\frac{x}{2}) = 0

4sin^2(x)+2cos^2(x)=3

4 sin^{2} (x) + 2 cos^{2} (x) = 3

cos^2(θ)-3sin(2θ)+sin^2(θ)+2sin^2(2θ)=0

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5cos(x)-10sin(x)cos(x)=0

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sin(2x+4)=cos(46)

sin (2 x + 4) = cos (4 6^{\circ})