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受欢迎的三角函数 >

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

解答

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

解答

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

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

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

展开

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

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

使用分配律:

a (b + c) = ab + a c

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

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

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

化简

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

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

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

2 \cdot 1 \cdot u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

2 \cdot 2 u^{2} u^{2}

数字相乘:

2 \cdot 2 = 4

= 4 u^{2} u^{2}

使用指数法则:

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

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

= 4 u^{2 + 2}

数字相加:

2 + 2 = 4

= 4 u^{4}

= 2 u^{2} + 4 u^{4}

= 2 u^{2} + 4 u^{4}

2 u^{2} + 4 u^{4} = 2

将

2

para o lado esquerdo

2 u^{2} + 4 u^{4} = 2

两边减去

2

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

化简

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

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

改写成标准形式

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

4 v^{2} + 2 v - 2 = 0

解

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

4 v^{2} + 2 v - 2 = 0

使用求根公式求解

4 v^{2} + 2 v - 2 = 0

二次方程求根公式:

若

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

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

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

2^{2} - 4 \cdot 4 (- 2) = 6

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

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

v_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

将解分隔开

v_{1} = \frac{- 2 + 6}{2 \cdot 4}, v_{2} = \frac{- 2 - 6}{2 \cdot 4}

v = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

\frac{- 2 + 6}{2 \cdot 4}

数字相加/相减:

- 2 + 6 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{8}

约分:

4

= \frac{1}{2}

v = \frac{- 2 - 6}{2 \cdot 4} : - 1

\frac{- 2 - 6}{2 \cdot 4}

数字相减:

- 2 - 6 = - 8

= \frac{- 8}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 8}{8}

使用分式法则:

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

= - \frac{8}{8}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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}

解

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

对于

x^{2} = f (a)

解为

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

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

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

u = cos (x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

cos (x) = i :

无解

cos (x) = i

无解

cos (x) = - i :

无解

cos (x) = - i

无解

合并所有解

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

以小数形式表示解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

作图

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