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受欢迎的三角函数 >

2cos^2(x)=11cos(x)+1

解答

2 cos^{2} (x) = 11 cos (x) + 1

解答

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

+1

度数

x = 95.13220 \dots^{\circ} + 36 0^{\circ} n, x = - 95.13220 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (x) = 11 cos (x) + 1

用替代法求解

2 cos^{2} (x) = 11 cos (x) + 1

令

: cos (x) = u

2 u^{2} = 11 u + 1

2 u^{2} = 11 u + 1 : u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

2 u^{2} = 11 u + 1

将

1

para o lado esquerdo

2 u^{2} = 11 u + 1

两边减去

1

2 u^{2} - 1 = 11 u + 1 - 1

化简

2 u^{2} - 1 = 11 u

2 u^{2} - 1 = 11 u

将

11 u

para o lado esquerdo

2 u^{2} - 1 = 11 u

两边减去

11 u

2 u^{2} - 1 - 11 u = 11 u - 11 u

化简

2 u^{2} - 1 - 11 u = 0

2 u^{2} - 1 - 11 u = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} - 11 u - 1 = 0

使用求根公式求解

2 u^{2} - 11 u - 1 = 0

二次方程求根公式:

若

a = 2, b = - 11, c = - 1

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 1) = 129

(- 11)^{2} - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= (- 11)^{2} + 4 \cdot 2 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 11)^{2} = 1 1^{2}

= 1 1^{2} + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 1^{2} + 8

1 1^{2} = 121

= 121 + 8

数字相加:

121 + 8 = 129

= 129

u_{1, 2} = \frac{- ( - 11 ) \pm 129}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 11 ) + 129}{2 \cdot 2}, u_{2} = \frac{- ( - 11 ) - 129}{2 \cdot 2}

u = \frac{- ( - 11 ) + 129}{2 \cdot 2} : \frac{11 + 129}{4}

\frac{- ( - 11 ) + 129}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{11 + 129}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{11 + 129}{4}

u = \frac{- ( - 11 ) - 129}{2 \cdot 2} : \frac{11 - 129}{4}

\frac{- ( - 11 ) - 129}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{11 - 129}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{11 - 129}{4}

二次方程组的解是:

u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

u = cos (x)

代回

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4} :

无解

cos (x) = \frac{11 + 129}{4}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{11 - 129}{4} : x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

cos (x) = \frac{11 - 129}{4}

使用反三角函数性质

cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 - 129}{4}

的通解

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

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

合并所有解

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

以小数形式表示解

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

作图

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