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受欢迎的三角函数 >

2tan^2(x)+1=cos(x)

解答

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

解答

x = 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边进行平方

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

两边减去

cos^{2} (x)

(2 tan^{2} (x) + 1)^{2} - cos^{2} (x) = 0

使用三角恒等式改写

(1 + 2 tan^{2} (x))^{2} - cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= (1 + 2 (sec^{2} (x) - 1))^{2} - cos^{2} (x)

乘开

1 + 2 (sec^{2} (x) - 1) : 2 sec^{2} (x) - 1

1 + 2 (sec^{2} (x) - 1)

乘开

2 (sec^{2} (x) - 1) : 2 sec^{2} (x) - 2

2 (sec^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = sec^{2} (x), c = 1

= 2 sec^{2} (x) - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 sec^{2} (x) - 2

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

化简

1 + 2 sec^{2} (x) - 2 : 2 sec^{2} (x) - 1

1 + 2 sec^{2} (x) - 2

对同类项分组

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

数字相加/相减:

1 - 2 = - 1

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

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

= (2 sec^{2} (x) - 1)^{2} - cos^{2} (x)

(- 1 + 2 sec^{2} (x))^{2} - cos^{2} (x) = 0

分解

(- 1 + 2 sec^{2} (x))^{2} - cos^{2} (x) : (- 1 + 2 sec^{2} (x) + cos (x)) (- 1 + 2 sec^{2} (x) - cos (x))

(- 1 + 2 sec^{2} (x))^{2} - cos^{2} (x)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(- 1 + 2 sec^{2} (x))^{2} - cos^{2} (x) = ((- 1 + 2 sec^{2} (x)) + cos (x)) ((- 1 + 2 sec^{2} (x)) - cos (x))

= ((- 1 + 2 sec^{2} (x)) + cos (x)) ((- 1 + 2 sec^{2} (x)) - cos (x))

整理后得

= (2 sec^{2} (x) + cos (x) - 1) (2 sec^{2} (x) - cos (x) - 1)

(- 1 + 2 sec^{2} (x) + cos (x)) (- 1 + 2 sec^{2} (x) - cos (x)) = 0

分别求解每个部分

- 1 + 2 sec^{2} (x) + cos (x) = 0 or - 1 + 2 sec^{2} (x) - cos (x) = 0

- 1 + 2 sec^{2} (x) + cos (x) = 0 : x = π + 2 πn

- 1 + 2 sec^{2} (x) + cos (x) = 0

使用三角恒等式改写

- 1 + cos (x) + 2 sec^{2} (x)

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

- 1 \cdot u + \frac{1}{u} u + 2 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{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

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

2 u^{2} u

使用指数法则:

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

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

= 2 u^{2 + 1}

数字相加:

2 + 1 = 3

= 2 u^{3}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- u + 1 + 2 u^{3} = 0

- u + 1 + 2 u^{3} = 0

- u + 1 + 2 u^{3} = 0

解

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

- u + 1 + 2 u^{3} = 0

改写成标准形式

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

2 u^{3} - u + 1 = 0

因式分解

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

2 u^{3} - u + 1

使用有理根定理

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

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2

因此，检验以下有理数:

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

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

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

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

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

对

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

做除法:

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

将分子

2 u^{3} - u + 1

与除数

u + 1

的首项系数相除：

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

商 = 2 u^{2}

将

u + 1

乘以

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

将

2 u^{3} - u + 1

减去

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

得到新的余数

余数 = - 2 u^{2} - u + 1

因此

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

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

对

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

做除法:

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

将分子

- 2 u^{2} - u + 1

与除数

u + 1

的首项系数相除：

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

商 = - 2 u

将

u + 1

乘以

- 2 u : - 2 u^{2} - 2 u

将

- 2 u^{2} - u + 1

减去

- 2 u^{2} - 2 u

得到新的余数

余数 = u + 1

因此

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

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

对

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

做除法:

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

将分子

u + 1

与除数

u + 1

的首项系数相除：

\frac{u}{u} = 1

商 = 1

将

u + 1

乘以

1 : u + 1

将

u + 1

减去

u + 1

得到新的余数

余数 = 0

因此

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

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

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

化简

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

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

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

使用虚数运算法则:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

数字相加/相减:

- 4 + 8 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 2 i}{4}

分解

2 + 2 i : 2 (1 + i)

2 + 2 i

改写为

= 2 \cdot 1 + 2 i

因式分解出通项

2

= 2 (1 + i)

= \frac{2 ( 1 + i )}{4}

约分:

2

= \frac{1 + i}{2}

将

\frac{1 + i}{2}

改写成标准复数形式:

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

\frac{1 + i}{2}

使用分式法则:

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

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

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 - 2 i}{4}

分解

2 - 2 i : 2 (1 - i)

2 - 2 i

改写为

= 2 \cdot 1 - 2 i

因式分解出通项

2

= 2 (1 - i)

= \frac{2 ( 1 - i )}{4}

约分:

2

= \frac{1 - i}{2}

将

\frac{1 - i}{2}

改写成标准复数形式:

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

\frac{1 - i}{2}

使用分式法则:

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

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

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

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

二次方程组的解是:

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

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

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

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

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

sec (x) = - 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = π + 2 πn

x = π + 2 πn

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

无解

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

无解

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

无解

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

无解

合并所有解

x = π + 2 πn

- 1 + 2 sec^{2} (x) - cos (x) = 0 : x = 2 πn

- 1 + 2 sec^{2} (x) - cos (x) = 0

使用三角恒等式改写

- 1 - cos (x) + 2 sec^{2} (x)

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

- 1 \cdot u - \frac{1}{u} u + 2 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{1}{u} u : - 1

- \frac{1}{u} u

分式相乘:

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

= - \frac{1 \cdot u}{u}

约分:

u

= - 1

化简

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

2 u^{2} u

使用指数法则:

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

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

= 2 u^{2 + 1}

数字相加:

2 + 1 = 3

= 2 u^{3}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- u - 1 + 2 u^{3} = 0

- u - 1 + 2 u^{3} = 0

- u - 1 + 2 u^{3} = 0

解

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

- u - 1 + 2 u^{3} = 0

改写成标准形式

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

2 u^{3} - u - 1 = 0

因式分解

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

2 u^{3} - u - 1

使用有理根定理

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

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2

因此，检验以下有理数:

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

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

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

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

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

对

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

做除法:

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

将分子

2 u^{3} - u - 1

与除数

u - 1

的首项系数相除：

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

商 = 2 u^{2}

将

u - 1

乘以

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

将

2 u^{3} - u - 1

减去

2 u^{3} - 2 u^{2}

得到新的余数

余数 = 2 u^{2} - u - 1

因此

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

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

对

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

做除法:

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

将分子

2 u^{2} - u - 1

与除数

u - 1

的首项系数相除：

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

商 = 2 u

将

u - 1

乘以

2 u : 2 u^{2} - 2 u

将

2 u^{2} - u - 1

减去

2 u^{2} - 2 u

得到新的余数

余数 = u - 1

因此

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

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

对

\frac{u - 1}{u - 1}

做除法:

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

将分子

u - 1

与除数

u - 1

的首项系数相除：

\frac{u}{u} = 1

商 = 1

将

u - 1

乘以

1 : u - 1

将

u - 1

减去

u - 1

得到新的余数

余数 = 0

因此

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

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

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 2, c = 1

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

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

化简

2^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

使用虚数运算法则:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

数字相加/相减:

- 4 + 8 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

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

将解分隔开

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

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

\frac{- 2 + 2 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 2 i}{4}

分解

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

改写为

= - 2 \cdot 1 + 2 i

因式分解出通项

2

= 2 (- 1 + i)

= \frac{2 ( - 1 + i )}{4}

约分:

2

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

将

\frac{- 1 + i}{2}

改写成标准复数形式:

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

\frac{- 1 + i}{2}

使用分式法则:

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

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

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

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

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

\frac{- 2 - 2 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 2 i}{4}

分解

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

改写为

= - 2 \cdot 1 - 2 i

因式分解出通项

2

= - 2 (1 + i)

= - \frac{2 ( 1 + i )}{4}

约分:

2

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

将

- \frac{1 + i}{2}

改写成标准复数形式:

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

- \frac{1 + i}{2}

使用分式法则:

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

\frac{1 + i}{2} = - (\frac{1}{2}) - (\frac{i}{2})

= - (\frac{1}{2}) - (\frac{i}{2})

去除括号:

(a) = a

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

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

二次方程组的解是:

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

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

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

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

sec (x) = 1 : x = 2 πn

sec (x) = 1

sec (x) = 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

sec (x) = - \frac{1}{2} + i \frac{1}{2} :

无解

sec (x) = - \frac{1}{2} + i \frac{1}{2}

无解

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

无解

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

无解

合并所有解

x = 2 πn

合并所有解

x = π + 2 πn, x = 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

2 tan^{2} (π + 2 π 1) + 1 = cos (π + 2 π 1)

整理后得

1 = - 1

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

2 tan^{2} (2 π 1) + 1 = cos (2 π 1)

整理后得

1 = 1

\Rightarrow 真

x = 2 πn

作图

流行的例子

sec (x) = - \frac{4}{3}

sin (B) = \frac{1}{4}

55sin(θ)-20-20cos(θ)=0

55 sin (θ) - 20 - 20 cos (θ) = 0

8 sin^{2} (x) = 1

5sin(x)cos(x)=cos(x)

5 sin (x) cos (x) = cos (x)