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受欢迎的三角函数 >

1.04cos^2(x)+0.102cos(x)-0.935=0

解答

1.04 cos^{2} (x) + 0.102 cos (x) - 0.935 = 0

解答

x = 0.45009 \dots + 2 πn, x = 2 π - 0.45009 \dots + 2 πn, x = 3.08648 \dots + 2 πn, x = - 3.08648 \dots + 2 πn

+1

度数

x = 25.78862 \dots^{\circ} + 36 0^{\circ} n, x = 334.21137 \dots^{\circ} + 36 0^{\circ} n, x = 176.84271 \dots^{\circ} + 36 0^{\circ} n, x = - 176.84271 \dots^{\circ} + 36 0^{\circ} n

求解步骤

1.04 cos^{2} (x) + 0.102 cos (x) - 0.935 = 0

用替代法求解

1.04 cos^{2} (x) + 0.102 cos (x) - 0.935 = 0

令

: cos (x) = u

1.04 u^{2} + 0.102 u - 0.935 = 0

1.04 u^{2} + 0.102 u - 0.935 = 0 : u = \frac{- 51 + 975001}{1040}, u = - \frac{51 + 975001}{1040}

1.04 u^{2} + 0.102 u - 0.935 = 0

在两边乘以

1000

1.04 u^{2} + 0.102 u - 0.935 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

3

digits to the right of the decimal point, therefore multiply by

1000

1.04 u^{2} \cdot 1000 + 0.102 u \cdot 1000 - 0.935 \cdot 1000 = 0 \cdot 1000

整理后得

1040 u^{2} + 102 u - 935 = 0

1040 u^{2} + 102 u - 935 = 0

使用求根公式求解

1040 u^{2} + 102 u - 935 = 0

二次方程求根公式:

若

a = 1040, b = 102, c = - 935

u_{1, 2} = \frac{- 102 \pm 10 2 ^{2} - 4 \cdot 1040 ( - 935 )}{2 \cdot 1040}

u_{1, 2} = \frac{- 102 \pm 10 2 ^{2} - 4 \cdot 1040 ( - 935 )}{2 \cdot 1040}

10 2^{2} - 4 \cdot 1040 (- 935) = 2975001

10 2^{2} - 4 \cdot 1040 (- 935)

使用法则

- (- a) = a

= 10 2^{2} + 4 \cdot 1040 \cdot 935

数字相乘:

4 \cdot 1040 \cdot 935 = 3889600

= 10 2^{2} + 3889600

10 2^{2} = 10404

= 10404 + 3889600

数字相加:

10404 + 3889600 = 3900004

= 3900004

3900004

质因数分解:

2^{2} \cdot 17 \cdot 83 \cdot 691

3900004

= 2^{2} \cdot 17 \cdot 83 \cdot 691

使用根式运算法则:

n ab = n a n b

= 2^{2} 17 \cdot 83 \cdot 691

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 17 \cdot 83 \cdot 691

整理后得

= 2975001

u_{1, 2} = \frac{- 102 \pm 2 975001}{2 \cdot 1040}

将解分隔开

u_{1} = \frac{- 102 + 2 975001}{2 \cdot 1040}, u_{2} = \frac{- 102 - 2 975001}{2 \cdot 1040}

u = \frac{- 102 + 2 975001}{2 \cdot 1040} : \frac{- 51 + 975001}{1040}

\frac{- 102 + 2 975001}{2 \cdot 1040}

数字相乘:

2 \cdot 1040 = 2080

= \frac{- 102 + 2 975001}{2080}

分解

- 102 + 2975001 : 2 (- 51 + 975001)

- 102 + 2975001

改写为

= - 2 \cdot 51 + 2975001

因式分解出通项

2

= 2 (- 51 + 975001)

= \frac{2 ( - 51 + 975001 )}{2080}

约分:

2

= \frac{- 51 + 975001}{1040}

u = \frac{- 102 - 2 975001}{2 \cdot 1040} : - \frac{51 + 975001}{1040}

\frac{- 102 - 2 975001}{2 \cdot 1040}

数字相乘:

2 \cdot 1040 = 2080

= \frac{- 102 - 2 975001}{2080}

分解

- 102 - 2975001 : - 2 (51 + 975001)

- 102 - 2975001

改写为

= - 2 \cdot 51 - 2975001

因式分解出通项

2

= - 2 (51 + 975001)

= - \frac{2 ( 51 + 975001 )}{2080}

约分:

2

= - \frac{51 + 975001}{1040}

二次方程组的解是:

u = \frac{- 51 + 975001}{1040}, u = - \frac{51 + 975001}{1040}

u = cos (x)

代回

cos (x) = \frac{- 51 + 975001}{1040}, cos (x) = - \frac{51 + 975001}{1040}

cos (x) = \frac{- 51 + 975001}{1040}, cos (x) = - \frac{51 + 975001}{1040}

cos (x) = \frac{- 51 + 975001}{1040} : x = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, x = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

cos (x) = \frac{- 51 + 975001}{1040}

使用反三角函数性质

cos (x) = \frac{- 51 + 975001}{1040}

cos (x) = \frac{- 51 + 975001}{1040}

的通解

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

x = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, x = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

x = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, x = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

cos (x) = - \frac{51 + 975001}{1040} : x = arccos (- \frac{51 + 975001}{1040}) + 2 πn, x = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

cos (x) = - \frac{51 + 975001}{1040}

使用反三角函数性质

cos (x) = - \frac{51 + 975001}{1040}

cos (x) = - \frac{51 + 975001}{1040}

的通解

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

x = arccos (- \frac{51 + 975001}{1040}) + 2 πn, x = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

x = arccos (- \frac{51 + 975001}{1040}) + 2 πn, x = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

合并所有解

x = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, x = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn, x = arccos (- \frac{51 + 975001}{1040}) + 2 πn, x = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

以小数形式表示解

x = 0.45009 \dots + 2 πn, x = 2 π - 0.45009 \dots + 2 πn, x = 3.08648 \dots + 2 πn, x = - 3.08648 \dots + 2 πn

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