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受欢迎的三角函数 >

sec(4x)-sec(2x)=2

解答

sec (4 x) - sec (2 x) = 2

解答

x = \frac{π}{2} + πn, x = \frac{0.62831 \dots}{2} + πn, x = π - \frac{0.62831 \dots}{2} + πn, x = \frac{1.88495 \dots}{2} + πn, x = - \frac{1.88495 \dots}{2} + πn

+1

度数

x = 9 0^{\circ} + 18 0^{\circ} n, x = 1 8^{\circ} + 18 0^{\circ} n, x = 16 2^{\circ} + 18 0^{\circ} n, x = 5 4^{\circ} + 18 0^{\circ} n, x = - 5 4^{\circ} + 18 0^{\circ} n

求解步骤

sec (4 x) - sec (2 x) = 2

两边减去

2

sec (4 x) - sec (2 x) - 2 = 0

使用三角恒等式改写

- 2 - sec (2 x) + sec (4 x)

使用基本三角恒等式:

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

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

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

改写为

= cos (2 \cdot 2 x)

使用倍角公式:

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

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

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

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

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

用替代法求解

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

令

: cos (2 x) = u

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

- 2 + \frac{1}{- 1 + 2 u ^{2}} - \frac{1}{u} = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

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

乘以最小公倍数

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

找到

- 1 + 2 u^{2}, u

的最小公倍数:

u (2 u + 1) (2 u - 1)

- 1 + 2 u^{2}, u

最小公倍数 (LCM)

因式分解表达式

分解

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

- 1 + 2 u^{2}

将

2 u^{2} - 1

改写为

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

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} u^{2} = (2 u)^{2}

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

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

使用平方差公式:

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

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

= (2 u + 1) (2 u - 1)

计算出由出现在

(2 u + 1) (2 u - 1)

或

u

中的因子组成的表达式

= u (2 u + 1) (2 u - 1)

乘以最小公倍数=

u (2 u + 1) (2 u - 1)

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

化简

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

化简

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

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

分式相乘:

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

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

乘以:

1 \cdot u = u

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

分解

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

2 u^{2} - 1

将

2 u^{2} - 1

改写为

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

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} u^{2} = (2 u)^{2}

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

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

使用平方差公式:

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

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

= (2 u + 1) (2 u - 1)

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

消掉

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

\frac{u ( 2 u + 1 ) ( 2 u - 1 )}{( 2 u + 1 ) ( 2 u - 1 )}

约分:

2 u + 1

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

约分:

2 u - 1

= u

= u

化简

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

- \frac{1}{u} u (2 u + 1) (2 u - 1)

分式相乘:

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

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

约分:

u

= - 1 \cdot (2 u + 1) (2 u - 1)

乘以:

1 \cdot (2 u + 1) = (2 u + 1)

= - (2 u + 1) (2 u - 1)

化简

0 \cdot u (2 u + 1) (2 u - 1) : 0

0 \cdot u (2 u + 1) (2 u - 1)

使用法则

0 \cdot a = 0

= 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

解

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) = 0

因式分解

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) : - (u + 1) (4 u^{2} - 2 u - 1)

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1)

乘开

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1) : - 4 u^{3} + 3 u - 2 u^{2} + 1

- 2 u (2 u + 1) (2 u - 1) + u - (2 u + 1) (2 u - 1)

乘开

- 2 u (2 u + 1) (2 u - 1) : - 4 u^{3} + 2 u

乘开

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

(2 u + 1) (2 u - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

(2 u)^{2} = 2 u^{2}

(2 u)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= (2)^{2} u^{2}

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 2 u^{2}

= 2 u^{2} - 1

= 2 u^{2} - 1

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

乘开

- 2 u (2 u^{2} - 1) : - 4 u^{3} + 2 u

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

使用分配律:

a (b - c) = ab - a c

a = - 2 u, b = 2 u^{2}, c = 1

= - 2 u \cdot 2 u^{2} - (- 2 u) \cdot 1

使用加减运算法则

- (- a) = a

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

化简

- 2 \cdot 2 u^{2} u + 2 \cdot 1 \cdot u : - 4 u^{3} + 2 u

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

2 \cdot 2 u^{2} u = 4 u^{3}

2 \cdot 2 u^{2} u

数字相乘:

2 \cdot 2 = 4

= 4 u^{2} u

使用指数法则:

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

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

= 4 u^{2 + 1}

数字相加:

2 + 1 = 3

= 4 u^{3}

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

数字相乘:

2 \cdot 1 = 2

= 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u

= - 4 u^{3} + 2 u + u - (2 u + 1) (2 u - 1)

乘开

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

乘开

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

(2 u + 1) (2 u - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

(2 u)^{2} = 2 u^{2}

(2 u)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= (2)^{2} u^{2}

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 2 u^{2}

= 2 u^{2} - 1

= 2 u^{2} - 1

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

= - 2 u^{2} + 1

= - 4 u^{3} + 2 u + u - 2 u^{2} + 1

同类项相加:

2 u + u = 3 u

= - 4 u^{3} + 3 u - 2 u^{2} + 1

= - 4 u^{3} + 3 u - 2 u^{2} + 1

分解

- 4 u^{3} - 2 u^{2} + 3 u + 1 : - (u + 1) (4 u^{2} - 2 u - 1)

- 4 u^{3} - 2 u^{2} + 3 u + 1

因式分解出通项

- 1

= - (4 u^{3} + 2 u^{2} - 3 u - 1)

分解

4 u^{3} + 2 u^{2} - 3 u - 1 : (u + 1) (4 u^{2} - 2 u - 1)

4 u^{3} + 2 u^{2} - 3 u - 1

使用有理根定理

a_{0} = 1, a_{n} = 4

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} - 2 u - 1

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

对

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

做除法:

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

将分子

4 u^{3} + 2 u^{2} - 3 u - 1

与除数

u + 1

的首项系数相除：

\frac{4 u ^{3}}{u} = 4 u^{2}

商 = 4 u^{2}

将

u + 1

乘以

4 u^{2} : 4 u^{3} + 4 u^{2}

将

4 u^{3} + 2 u^{2} - 3 u - 1

减去

4 u^{3} + 4 u^{2}

得到新的余数

余数 = - 2 u^{2} - 3 u - 1

因此

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

= 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

对

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

做除法:

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

将分子

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

= 4 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

= 4 u^{2} - 2 u - 1

= 4 u^{2} - 2 u - 1

= (u + 1) (4 u^{2} - 2 u - 1)

= - (u + 1) (4 u^{2} - 2 u - 1)

= - (u + 1) (4 u^{2} - 2 u - 1)

- (u + 1) (4 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 4 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

4 u^{2} - 2 u - 1 = 0 : u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 4} : \frac{1 + 5}{4}

\frac{- ( - 2 ) + 2 5}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 + 2 5}{8}

分解

2 + 25 : 2 (1 + 5)

2 + 25

改写为

= 2 \cdot 1 + 25

因式分解出通项

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

约分:

2

= \frac{1 + 5}{4}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 4} : \frac{1 - 5}{4}

\frac{- ( - 2 ) - 2 5}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 - 2 5}{8}

分解

2 - 25 : 2 (1 - 5)

2 - 25

改写为

= 2 \cdot 1 - 25

因式分解出通项

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{8}

约分:

2

= \frac{1 - 5}{4}

二次方程组的解是:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

解为

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

验证解

找到无定义的点（奇点）:

u = \frac{1}{2}, u = - \frac{1}{2}, u = 0

取

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

的分母，令其等于零

解

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

- 1 + 2 u^{2} = 0

将

1

到右边

- 1 + 2 u^{2} = 0

两边加上

1

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

化简

2 u^{2} = 1

2 u^{2} = 1

两边除以

2

2 u^{2} = 1

两边除以

2

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

化简

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

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

对于

x^{2} = f (a)

解为

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

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

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

\frac{1}{2}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{2}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{2}

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

- \frac{1}{2}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{2}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{2}

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

u = 0

以下点无定义

u = \frac{1}{2}, u = - \frac{1}{2}, u = 0

将不在定义域的点与解相综合:

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

u = cos (2 x)

代回

cos (2 x) = - 1, cos (2 x) = \frac{1 + 5}{4}, cos (2 x) = \frac{1 - 5}{4}

cos (2 x) = - 1, cos (2 x) = \frac{1 + 5}{4}, cos (2 x) = \frac{1 - 5}{4}

cos (2 x) = - 1 : x = \frac{π}{2} + πn

cos (2 x) = - 1

cos (2 x) = - 1

的通解

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

2 x = π + 2 πn

2 x = π + 2 πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

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

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cos (2 x) = \frac{1 + 5}{4} : x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

cos (2 x) = \frac{1 + 5}{4}

使用反三角函数性质

cos (2 x) = \frac{1 + 5}{4}

cos (2 x) = \frac{1 + 5}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn, 2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn, 2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

解

2 x = arccos (\frac{1 + 5}{4}) + 2 πn : x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

2 x = arccos (\frac{1 + 5}{4}) + 2 πn

两边除以

2

2 x = arccos (\frac{1 + 5}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{1 + 5}{4} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

解

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn : x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

两边除以

2

2 x = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1 + 5}{4} )}{2} + \frac{2 πn}{2}

化简

x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn

cos (2 x) = \frac{1 - 5}{4} : x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

cos (2 x) = \frac{1 - 5}{4}

使用反三角函数性质

cos (2 x) = \frac{1 - 5}{4}

cos (2 x) = \frac{1 - 5}{4}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn, 2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn, 2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

解

2 x = arccos (\frac{1 - 5}{4}) + 2 πn : x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

2 x = arccos (\frac{1 - 5}{4}) + 2 πn

两边除以

2

2 x = arccos (\frac{1 - 5}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{1 - 5}{4} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

解

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn : x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

两边除以

2

2 x = - arccos (\frac{1 - 5}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

合并所有解

x = \frac{π}{2} + πn, x = \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1 + 5}{4} )}{2} + πn, x = \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn, x = - \frac{arccos ( \frac{1 - 5}{4} )}{2} + πn

以小数形式表示解

x = \frac{π}{2} + πn, x = \frac{0.62831 \dots}{2} + πn, x = π - \frac{0.62831 \dots}{2} + πn, x = \frac{1.88495 \dots}{2} + πn, x = - \frac{1.88495 \dots}{2} + πn

作图

流行的例子

- cos (x) = 1

cot(x)sin(x)-sin(x)=0

cot (x) sin (x) - sin (x) = 0

1.16cos(θ)+0.532cos(θ)-0.557=0

1.16 cos (θ) + 0.532 cos (θ) - 0.557 = 0

arcsin(6x)+arcsin(6sqrt(3)x)=-pi/2

arcsin (6 x) + arcsin (63 x) = - \frac{π}{2}

cos(x)= 1/2 ,0<= x<= 360

cos (x) = \frac{1}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}