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受欢迎的三角函数 >

cos^2(2x+pi/6)= 1/2

解答

cos^{2} (2 x + \frac{π}{6}) = \frac{1}{2}

解答

x = πn + \frac{π}{24}, x = π + πn - \frac{5 π}{24}, x = πn + \frac{7 π}{24}, x = πn - \frac{11 π}{24}

+1

度数

x = 7. 5^{\circ} + 18 0^{\circ} n, x = 142. 5^{\circ} + 18 0^{\circ} n, x = 52. 5^{\circ} + 18 0^{\circ} n, x = - 82. 5^{\circ} + 18 0^{\circ} n

求解步骤

cos^{2} (2 x + \frac{π}{6}) = \frac{1}{2}

用替代法求解

cos^{2} (2 x + \frac{π}{6}) = \frac{1}{2}

令

: cos (2 x + \frac{π}{6}) = u

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

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

u = cos (2 x + \frac{π}{6})

代回

cos (2 x + \frac{π}{6}) = \frac{1}{2}, cos (2 x + \frac{π}{6}) = - \frac{1}{2}

cos (2 x + \frac{π}{6}) = \frac{1}{2}, cos (2 x + \frac{π}{6}) = - \frac{1}{2}

cos (2 x + \frac{π}{6}) = \frac{1}{2} : x = πn + \frac{π}{24}, x = π + πn - \frac{5 π}{24}

cos (2 x + \frac{π}{6}) = \frac{1}{2}

使用反三角函数性质

cos (2 x + \frac{π}{6}) = \frac{1}{2}

cos (2 x + \frac{π}{6}) = \frac{1}{2}

的通解

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

2 x + \frac{π}{6} = arccos (\frac{1}{2}) + 2 πn, 2 x + \frac{π}{6} = 2 π - arccos (\frac{1}{2}) + 2 πn

2 x + \frac{π}{6} = arccos (\frac{1}{2}) + 2 πn, 2 x + \frac{π}{6} = 2 π - arccos (\frac{1}{2}) + 2 πn

解

2 x + \frac{π}{6} = arccos (\frac{1}{2}) + 2 πn : x = πn + \frac{π}{24}

2 x + \frac{π}{6} = arccos (\frac{1}{2}) + 2 πn

化简

arccos (\frac{1}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

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

将

\frac{π}{6}

到右边

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

两边减去

\frac{π}{6}

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

化简

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

化简

2 x + \frac{π}{6} - \frac{π}{6} : 2 x

2 x + \frac{π}{6} - \frac{π}{6}

同类项相加:

\frac{π}{6} - \frac{π}{6} = 0

= 2 x

化简

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

\frac{π}{4} + 2 πn - \frac{π}{6}

对同类项分组

= 2 πn + \frac{π}{4} - \frac{π}{6}

4, 6

的最小公倍数:

12

4, 6

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

4

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{π}{6} :

将分母和分子乘以

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

= \frac{π 3}{12} - \frac{π 2}{12}

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 3 - π 2}{12}

同类项相加:

3 π - 2 π = π

= 2 πn + \frac{π}{12}

2 x = 2 πn + \frac{π}{12}

2 x = 2 πn + \frac{π}{12}

2 x = 2 πn + \frac{π}{12}

两边除以

2

2 x = 2 πn + \frac{π}{12}

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{12}}{2} = \frac{π}{24}

\frac{\frac{π}{12}}{2}

使用分式法则:

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

= \frac{π}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= \frac{π}{24}

= πn + \frac{π}{24}

x = πn + \frac{π}{24}

x = πn + \frac{π}{24}

x = πn + \frac{π}{24}

解

2 x + \frac{π}{6} = 2 π - arccos (\frac{1}{2}) + 2 πn : x = π + πn - \frac{5 π}{24}

2 x + \frac{π}{6} = 2 π - arccos (\frac{1}{2}) + 2 πn

化简

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

2 π - arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{4} + 2 πn

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

将

\frac{π}{6}

到右边

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

两边减去

\frac{π}{6}

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

化简

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

化简

2 x + \frac{π}{6} - \frac{π}{6} : 2 x

2 x + \frac{π}{6} - \frac{π}{6}

同类项相加:

\frac{π}{6} - \frac{π}{6} = 0

= 2 x

化简

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

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

对同类项分组

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

4, 6

的最小公倍数:

12

4, 6

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

4

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{π}{6} :

将分母和分子乘以

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

= - \frac{π 3}{12} - \frac{π 2}{12}

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 3 - π 2}{12}

同类项相加:

- 3 π - 2 π = - 5 π

= \frac{- 5 π}{12}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= 2 π + 2 πn - \frac{5 π}{12}

2 x = 2 π + 2 πn - \frac{5 π}{12}

2 x = 2 π + 2 πn - \frac{5 π}{12}

2 x = 2 π + 2 πn - \frac{5 π}{12}

两边除以

2

2 x = 2 π + 2 πn - \frac{5 π}{12}

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{2 π}{2} + \frac{2 πn}{2} - \frac{\frac{5 π}{12}}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{5 π}{12}}{2} = \frac{5 π}{24}

\frac{\frac{5 π}{12}}{2}

使用分式法则:

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

= \frac{5 π}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= \frac{5 π}{24}

= π + πn - \frac{5 π}{24}

x = π + πn - \frac{5 π}{24}

x = π + πn - \frac{5 π}{24}

x = π + πn - \frac{5 π}{24}

x = πn + \frac{π}{24}, x = π + πn - \frac{5 π}{24}

cos (2 x + \frac{π}{6}) = - \frac{1}{2} : x = πn + \frac{7 π}{24}, x = πn - \frac{11 π}{24}

cos (2 x + \frac{π}{6}) = - \frac{1}{2}

使用反三角函数性质

cos (2 x + \frac{π}{6}) = - \frac{1}{2}

cos (2 x + \frac{π}{6}) = - \frac{1}{2}

的通解

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

2 x + \frac{π}{6} = arccos (- \frac{1}{2}) + 2 πn, 2 x + \frac{π}{6} = - arccos (- \frac{1}{2}) + 2 πn

2 x + \frac{π}{6} = arccos (- \frac{1}{2}) + 2 πn, 2 x + \frac{π}{6} = - arccos (- \frac{1}{2}) + 2 πn

解

2 x + \frac{π}{6} = arccos (- \frac{1}{2}) + 2 πn : x = πn + \frac{7 π}{24}

2 x + \frac{π}{6} = arccos (- \frac{1}{2}) + 2 πn

化简

arccos (- \frac{1}{2}) + 2 πn : \frac{3 π}{4} + 2 πn

arccos (- \frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{3 π}{4} + 2 πn

2 x + \frac{π}{6} = \frac{3 π}{4} + 2 πn

将

\frac{π}{6}

到右边

2 x + \frac{π}{6} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{6}

2 x + \frac{π}{6} - \frac{π}{6} = \frac{3 π}{4} + 2 πn - \frac{π}{6}

化简

2 x + \frac{π}{6} - \frac{π}{6} = \frac{3 π}{4} + 2 πn - \frac{π}{6}

化简

2 x + \frac{π}{6} - \frac{π}{6} : 2 x

2 x + \frac{π}{6} - \frac{π}{6}

同类项相加:

\frac{π}{6} - \frac{π}{6} = 0

= 2 x

化简

\frac{3 π}{4} + 2 πn - \frac{π}{6} : 2 πn + \frac{7 π}{12}

\frac{3 π}{4} + 2 πn - \frac{π}{6}

对同类项分组

= 2 πn - \frac{π}{6} + \frac{3 π}{4}

6, 4

的最小公倍数:

12

6, 4

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

6

或

4

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{6} :

将分母和分子乘以

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

对于

\frac{3 π}{4} :

将分母和分子乘以

3

\frac{3 π}{4} = \frac{3 π 3}{4 \cdot 3} = \frac{9 π}{12}

= - \frac{π 2}{12} + \frac{9 π}{12}

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 + 9 π}{12}

同类项相加:

- 2 π + 9 π = 7 π

= 2 πn + \frac{7 π}{12}

2 x = 2 πn + \frac{7 π}{12}

2 x = 2 πn + \frac{7 π}{12}

2 x = 2 πn + \frac{7 π}{12}

两边除以

2

2 x = 2 πn + \frac{7 π}{12}

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{7 π}{12}}{2}

化简

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{7 π}{12}}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 πn}{2} + \frac{\frac{7 π}{12}}{2} : πn + \frac{7 π}{24}

\frac{2 πn}{2} + \frac{\frac{7 π}{12}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{7 π}{12}}{2} = \frac{7 π}{24}

\frac{\frac{7 π}{12}}{2}

使用分式法则:

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

= \frac{7 π}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= \frac{7 π}{24}

= πn + \frac{7 π}{24}

x = πn + \frac{7 π}{24}

x = πn + \frac{7 π}{24}

x = πn + \frac{7 π}{24}

解

2 x + \frac{π}{6} = - arccos (- \frac{1}{2}) + 2 πn : x = πn - \frac{11 π}{24}

2 x + \frac{π}{6} = - arccos (- \frac{1}{2}) + 2 πn

化简

- arccos (- \frac{1}{2}) + 2 πn : - \frac{3 π}{4} + 2 πn

- arccos (- \frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{3 π}{4} + 2 πn

2 x + \frac{π}{6} = - \frac{3 π}{4} + 2 πn

将

\frac{π}{6}

到右边

2 x + \frac{π}{6} = - \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{6}

2 x + \frac{π}{6} - \frac{π}{6} = - \frac{3 π}{4} + 2 πn - \frac{π}{6}

化简

2 x + \frac{π}{6} - \frac{π}{6} = - \frac{3 π}{4} + 2 πn - \frac{π}{6}

化简

2 x + \frac{π}{6} - \frac{π}{6} : 2 x

2 x + \frac{π}{6} - \frac{π}{6}

同类项相加:

\frac{π}{6} - \frac{π}{6} = 0

= 2 x

化简

- \frac{3 π}{4} + 2 πn - \frac{π}{6} : 2 πn - \frac{11 π}{12}

- \frac{3 π}{4} + 2 πn - \frac{π}{6}

对同类项分组

= 2 πn - \frac{π}{6} - \frac{3 π}{4}

6, 4

的最小公倍数:

12

6, 4

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

6

或

4

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{6} :

将分母和分子乘以

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

对于

\frac{3 π}{4} :

将分母和分子乘以

3

\frac{3 π}{4} = \frac{3 π 3}{4 \cdot 3} = \frac{9 π}{12}

= - \frac{π 2}{12} - \frac{9 π}{12}

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 - 9 π}{12}

同类项相加:

- 2 π - 9 π = - 11 π

= \frac{- 11 π}{12}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= 2 πn - \frac{11 π}{12}

2 x = 2 πn - \frac{11 π}{12}

2 x = 2 πn - \frac{11 π}{12}

2 x = 2 πn - \frac{11 π}{12}

两边除以

2

2 x = 2 πn - \frac{11 π}{12}

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2} - \frac{\frac{11 π}{12}}{2}

化简

\frac{2 x}{2} = \frac{2 πn}{2} - \frac{\frac{11 π}{12}}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 πn}{2} - \frac{\frac{11 π}{12}}{2} : πn - \frac{11 π}{24}

\frac{2 πn}{2} - \frac{\frac{11 π}{12}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{11 π}{12}}{2} = \frac{11 π}{24}

\frac{\frac{11 π}{12}}{2}

使用分式法则:

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

= \frac{11 π}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= \frac{11 π}{24}

= πn - \frac{11 π}{24}

x = πn - \frac{11 π}{24}

x = πn - \frac{11 π}{24}

x = πn - \frac{11 π}{24}

x = πn + \frac{7 π}{24}, x = πn - \frac{11 π}{24}

合并所有解

x = πn + \frac{π}{24}, x = π + πn - \frac{5 π}{24}, x = πn + \frac{7 π}{24}, x = πn - \frac{11 π}{24}

作图

流行的例子

cos(5x)=1,0<= x<= 2pi

cos (5 x) = 1, 0 \leq x \leq 2 π

sec(2θ)=2,0<θ< pi/2

sec (2 θ) = 2, 0 < θ < \frac{π}{2}

2 cos (2 θ) + 1 = 0

sec (\frac{5 θ}{4}) = 2

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

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