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受欢迎的三角函数 >

1-2cos^2(8x)=sin(4x)

解答

1 - 2 cos^{2} (8 x) = sin (4 x)

解答

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

+1

度数

x = 67. 5^{\circ} + 9 0^{\circ} n, x = 7. 5^{\circ} + 9 0^{\circ} n, x = 37. 5^{\circ} + 9 0^{\circ} n, x = 13. 5^{\circ} + 9 0^{\circ} n, x = 31. 5^{\circ} + 9 0^{\circ} n, x = - 4. 5^{\circ} + 9 0^{\circ} n, x = 49. 5^{\circ} + 9 0^{\circ} n

求解步骤

1 - 2 cos^{2} (8 x) = sin (4 x)

两边减去

sin (4 x)

1 - 2 cos^{2} (8 x) - sin (4 x) = 0

令

: u = 4 x

1 - 2 cos^{2} (2 u) - sin (u) = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

1 - sin (u) - 2 (1 - 2 sin^{2} (u))^{2} : 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

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

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

使用完全平方公式:

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

a = 1, b = 2 sin^{2} (u)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 2 sin^{2} (u) = 4 sin^{2} (u)

2 \cdot 1 \cdot 2 sin^{2} (u)

数字相乘:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (u)

(2 sin^{2} (u))^{2} = 4 sin^{4} (u)

(2 sin^{2} (u))^{2}

使用指数法则:

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

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

(sin^{2} (u))^{2} : sin^{4} (u)

使用指数法则:

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

= sin^{2 \cdot 2} (u)

数字相乘:

2 \cdot 2 = 4

= sin^{4} (u)

= 2^{2} sin^{4} (u)

2^{2} = 4

= 4 sin^{4} (u)

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

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

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

乘开

- 2 (1 - 4 sin^{2} (u) + 4 sin^{4} (u)) : - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

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

打开括号

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

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

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

化简

- 2 \cdot 1 + 2 \cdot 4 sin^{2} (u) - 2 \cdot 4 sin^{4} (u) : - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

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

数字相乘:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 4 = 8

= - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

= - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

= 1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

化简

1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u) : 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

1 - sin (u) - 2 + 8 sin^{2} (u) - 8 sin^{4} (u)

对同类项分组

= - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) + 1 - 2

数字相加/相减:

1 - 2 = - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

= 8 sin^{2} (u) - 8 sin^{4} (u) - sin (u) - 1

- 1 - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) = 0

用替代法求解

- 1 - sin (u) + 8 sin^{2} (u) - 8 sin^{4} (u) = 0

令

: sin (u) = u

- 1 - u + 8 u^{2} - 8 u^{4} = 0

- 1 - u + 8 u^{2} - 8 u^{4} = 0 : u = - 1, u = \frac{1}{2}, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

- 1 - u + 8 u^{2} - 8 u^{4} = 0

改写成标准形式

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

- 8 u^{4} + 8 u^{2} - u - 1 = 0

因式分解

- 8 u^{4} + 8 u^{2} - u - 1 : - (u + 1) (2 u - 1) (4 u^{2} - 2 u - 1)

- 8 u^{4} + 8 u^{2} - u - 1

因式分解出通项

- 1

= - (8 u^{4} - 8 u^{2} + u + 1)

分解

8 u^{4} - 8 u^{2} + u + 1 : (u + 1) (2 u - 1) (4 u^{2} - 2 u - 1)

8 u^{4} - 8 u^{2} + u + 1

使用有理根定理

a_{0} = 1, a_{n} = 8

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4, 8

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4 , 8}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1}

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} - 8 u^{2} + 1

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1}

对

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1}

做除法:

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

将分子

8 u^{4} - 8 u^{2} + u + 1

与除数

u + 1

的首项系数相除：

\frac{8 u ^{4}}{u} = 8 u^{3}

商 = 8 u^{3}

将

u + 1

乘以

8 u^{3} : 8 u^{4} + 8 u^{3}

将

8 u^{4} - 8 u^{2} + u + 1

减去

8 u^{4} + 8 u^{3}

得到新的余数

余数 = - 8 u^{3} - 8 u^{2} + u + 1

因此

\frac{8 u ^{4} - 8 u ^{2} + u + 1}{u + 1} = 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

= 8 u^{3} + \frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

对

\frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1}

做除法:

\frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1} = - 8 u^{2} + \frac{u + 1}{u + 1}

将分子

- 8 u^{3} - 8 u^{2} + u + 1

与除数

u + 1

的首项系数相除：

\frac{- 8 u ^{3}}{u} = - 8 u^{2}

商 = - 8 u^{2}

将

u + 1

乘以

- 8 u^{2} : - 8 u^{3} - 8 u^{2}

将

- 8 u^{3} - 8 u^{2} + u + 1

减去

- 8 u^{3} - 8 u^{2}

得到新的余数

余数 = u + 1

因此

\frac{- 8 u ^{3} - 8 u ^{2} + u + 1}{u + 1} = - 8 u^{2} + \frac{u + 1}{u + 1}

= 8 u^{3} - 8 u^{2} + \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

= 8 u^{3} - 8 u^{2} + 1

= 8 u^{3} - 8 u^{2} + 1

分解

8 u^{3} - 8 u^{2} + 1 : (2 u - 1) (4 u^{2} - 2 u - 1)

8 u^{3} - 8 u^{2} + 1

使用有理根定理

a_{0} = 1, a_{n} = 8

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4, 8

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4 , 8}

\frac{1}{2}

是表达式的根，所以因式分解

2 u - 1

= (2 u - 1) \frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1}

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} - 2 u - 1

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1}

对

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1}

做除法:

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} + \frac{- 4 u ^{2} + 1}{2 u - 1}

将分子

8 u^{3} - 8 u^{2} + 1

与除数

2 u - 1

的首项系数相除：

\frac{8 u ^{3}}{2 u} = 4 u^{2}

商 = 4 u^{2}

将

2 u - 1

乘以

4 u^{2} : 8 u^{3} - 4 u^{2}

将

8 u^{3} - 8 u^{2} + 1

减去

8 u^{3} - 4 u^{2}

得到新的余数

余数 = - 4 u^{2} + 1

因此

\frac{8 u ^{3} - 8 u ^{2} + 1}{2 u - 1} = 4 u^{2} + \frac{- 4 u ^{2} + 1}{2 u - 1}

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

对

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

做除法:

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

将分子

- 4 u^{2} + 1

与除数

2 u - 1

的首项系数相除：

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

商 = - 2 u

将

2 u - 1

乘以

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

将

- 4 u^{2} + 1

减去

- 4 u^{2} + 2 u

得到新的余数

余数 = - 2 u + 1

因此

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

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

对

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

做除法:

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

将分子

- 2 u + 1

与除数

2 u - 1

的首项系数相除：

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

商 = - 1

将

2 u - 1

乘以

- 1 : - 2 u + 1

将

- 2 u + 1

减去

- 2 u + 1

得到新的余数

余数 = 0

因此

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

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

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

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

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

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

- (u + 1) (2 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 2 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

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

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

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

化简

u = \frac{1}{2}

u = \frac{1}{2}

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

u = sin (u)

代回

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

sin (u) = - 1

sin (u) = - 1

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}

u = \frac{3 π}{2} + 2 πn

u = \frac{3 π}{2} + 2 πn

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

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

sin (u) = \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}

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

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

sin (u) = \frac{1 + 5}{4} : u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (u) = \frac{1 + 5}{4}

使用反三角函数性质

sin (u) = \frac{1 + 5}{4}

sin (u) = \frac{1 + 5}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

sin (u) = \frac{1 - 5}{4} : u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

sin (u) = \frac{1 - 5}{4}

使用反三角函数性质

sin (u) = \frac{1 - 5}{4}

sin (u) = \frac{1 - 5}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

合并所有解

u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = arcsin (\frac{1 + 5}{4}) + 2 πn, u = π - arcsin (\frac{1 + 5}{4}) + 2 πn, u = arcsin (\frac{1 - 5}{4}) + 2 πn, u = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

u = 4 x

代回

4 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{8}

4 x = \frac{3 π}{2} + 2 πn

两边除以

4

4 x = \frac{3 π}{2} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π + 4 πn}{8}

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

使用法则

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

= \frac{\frac{3 π}{2} + 2 πn}{4}

化简

\frac{3 π}{2} + 2 πn : \frac{3 π + 4 πn}{2}

\frac{3 π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{3 π}{2} + \frac{2 πn \cdot 2}{2}

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{3 π + 2 πn \cdot 2}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{2}

= \frac{\frac{3 π + 4 πn}{2}}{4}

使用分式法则:

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

= \frac{3 π + 4 πn}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

4 x = \frac{π}{6} + 2 πn : x = \frac{π + 12 πn}{24}

4 x = \frac{π}{6} + 2 πn

两边除以

4

4 x = \frac{π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π + 12 πn}{24}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

使用法则

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

= \frac{\frac{π}{6} + 2 πn}{4}

化简

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

\frac{π}{6} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 6}{6}

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

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{π + 2 πn \cdot 6}{6}

数字相乘:

2 \cdot 6 = 12

= \frac{π + 12 πn}{6}

= \frac{\frac{π + 12 πn}{6}}{4}

使用分式法则:

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

= \frac{π + 12 πn}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

x = \frac{π + 12 πn}{24}

4 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π + 12 πn}{24}

4 x = \frac{5 π}{6} + 2 πn

两边除以

4

4 x = \frac{5 π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4} : \frac{5 π + 12 πn}{24}

\frac{\frac{5 π}{6}}{4} + \frac{2 πn}{4}

使用法则

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

= \frac{\frac{5 π}{6} + 2 πn}{4}

化简

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

\frac{5 π}{6} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 6}{6}

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

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{5 π + 2 πn \cdot 6}{6}

数字相乘:

2 \cdot 6 = 12

= \frac{5 π + 12 πn}{6}

= \frac{\frac{5 π + 12 πn}{6}}{4}

使用分式法则:

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

= \frac{5 π + 12 πn}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

x = \frac{5 π + 12 πn}{24}

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

4 x = arcsin (\frac{1 + 5}{4}) + 2 πn

两边除以

4

4 x = arcsin (\frac{1 + 5}{4}) + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4} : \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

\frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4}

使用法则

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

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

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

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

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

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

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

两边除以

4

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

两边除以

4

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

化简

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

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{π}{4} - \frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4} : \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}

\frac{π}{4} - \frac{arcsin ( \frac{1 + 5}{4} )}{4} + \frac{2 πn}{4}

使用法则

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

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

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

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

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

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

4 x = arcsin (\frac{1 - 5}{4}) + 2 πn

两边除以

4

4 x = arcsin (\frac{1 - 5}{4}) + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{arcsin ( \frac{1 - 5}{4} )}{4} + \frac{2 πn}{4} : \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}

\frac{arcsin ( \frac{1 - 5}{4} )}{4} + \frac{2 πn}{4}

使用法则

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

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

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

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

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

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

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

- \frac{1 - 5}{4} = - \frac{- ( 5 - 1 )}{4} = \frac{5 - 1}{4}

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

两边除以

4

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

两边除以

4

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

化简

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

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{π}{4} + \frac{arcsin ( \frac{5 - 1}{4} )}{4} + \frac{2 πn}{4} : \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

\frac{π}{4} + \frac{arcsin ( \frac{5 - 1}{4} )}{4} + \frac{2 πn}{4}

使用法则

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

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

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

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

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

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{π - arcsin ( \frac{1 + 5}{4} ) + 2 πn}{4}, x = \frac{arcsin ( \frac{1 - 5}{4} ) + 2 πn}{4}, x = \frac{π + arcsin ( \frac{5 - 1}{4} ) + 2 πn}{4}

以小数形式表示解

x = \frac{3 π + 4 πn}{8}, x = \frac{π + 12 πn}{24}, x = \frac{5 π + 12 πn}{24}, x = \frac{0.94247 \dots + 2 πn}{4}, x = \frac{π - 0.94247 \dots + 2 πn}{4}, x = \frac{- 0.31415 \dots + 2 πn}{4}, x = \frac{π + 0.31415 \dots + 2 πn}{4}

作图

流行的例子

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sin(y)=(50)/(65.3)

sin (y) = \frac{50}{65.3}

0.26=(1-sin(x))/(1+sin(x))

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

sec(x)=-2,0<= x<= 2pi

sec (x) = - 2, 0 \leq x \leq 2 π