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受欢迎的三角函数 >

cos^4(x)= 1/2

解答

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

解答

x = 0.57185 \dots + 2 πn, x = 2 π - 0.57185 \dots + 2 πn, x = 2.56973 \dots + 2 πn, x = - 2.56973 \dots + 2 πn

+1

度数

x = 32.76509 \dots^{\circ} + 36 0^{\circ} n, x = 327.23490 \dots^{\circ} + 36 0^{\circ} n, x = 147.23490 \dots^{\circ} + 36 0^{\circ} n, x = - 147.23490 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

u^{4} = \frac{1}{2} : u = 4 \frac{1}{2}, u = - 4 \frac{1}{2}, u = i 4 \frac{1}{2}, u = - i 4 \frac{1}{2}

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

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

解

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

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

对于

(g (x))^{2} = f (a)

解为

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

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

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

代回

v = u^{2}

，求解

u

解

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

\frac{1}{2}

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= (\frac{1}{2})^{\frac{1}{4}}

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

= 4 \frac{1}{2}

- \frac{1}{2} = - 4 \frac{1}{2}

- \frac{1}{2}

化简

\frac{1}{2} : 4 \frac{1}{2}

\frac{1}{2}

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= (\frac{1}{2})^{\frac{1}{4}}

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

= 4 \frac{1}{2}

= - 4 \frac{1}{2}

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

解

u^{2} = - \frac{1}{2} : u = i 4 \frac{1}{2}, u = - i 4 \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}

化简

- \frac{1}{2} : i 4 \frac{1}{2}

- \frac{1}{2}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1}{2}

\frac{1}{2} : 4 \frac{1}{2}

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= (\frac{1}{2})^{\frac{1}{4}}

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

= 4 \frac{1}{2}

= i 4 \frac{1}{2}

化简

- - \frac{1}{2} : - i 4 \frac{1}{2}

- - \frac{1}{2}

化简

- \frac{1}{2} : i 4 \frac{1}{2}

- \frac{1}{2}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1}{2}

\frac{1}{2} : 4 \frac{1}{2}

使用根式运算法则:

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

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

使用指数法则:

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

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

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= (\frac{1}{2})^{\frac{1}{4}}

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

= 4 \frac{1}{2}

= i 4 \frac{1}{2}

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

u = i 4 \frac{1}{2}, u = - i 4 \frac{1}{2}

解为

u = 4 \frac{1}{2}, u = - 4 \frac{1}{2}, u = i 4 \frac{1}{2}, u = - i 4 \frac{1}{2}

u = cos (x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

cos (x) = i 4 \frac{1}{2} :

无解

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

无解

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

无解

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

无解

合并所有解

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

以小数形式表示解

x = 0.57185 \dots + 2 πn, x = 2 π - 0.57185 \dots + 2 πn, x = 2.56973 \dots + 2 πn, x = - 2.56973 \dots + 2 πn

作图

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