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受欢迎的三角函数 >

3sin(x)+6cos(x)=4

解答

3 sin (x) + 6 cos (x) = 4

解答

x = 1.39557 \dots + 2 πn, x = 2 π - 0.46828 \dots + 2 πn

+1

度数

x = 79.96077 \dots^{\circ} + 36 0^{\circ} n, x = 333.16932 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (x) + 6 cos (x) = 4

两边减去

6 cos (x)

3 sin (x) = 4 - 6 cos (x)

两边进行平方

(3 sin (x))^{2} = (4 - 6 cos (x))^{2}

两边减去

(4 - 6 cos (x))^{2}

9 sin^{2} (x) - 16 + 48 cos (x) - 36 cos^{2} (x) = 0

使用三角恒等式改写

- 16 - 36 cos^{2} (x) + 48 cos (x) + 9 sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 16 - 36 cos^{2} (x) + 48 cos (x) + 9 (1 - cos^{2} (x))

化简

- 16 - 36 cos^{2} (x) + 48 cos (x) + 9 (1 - cos^{2} (x)) : 48 cos (x) - 45 cos^{2} (x) - 7

- 16 - 36 cos^{2} (x) + 48 cos (x) + 9 (1 - cos^{2} (x))

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= - 16 - 36 cos^{2} (x) + 48 cos (x) + 9 - 9 cos^{2} (x)

化简

- 16 - 36 cos^{2} (x) + 48 cos (x) + 9 - 9 cos^{2} (x) : 48 cos (x) - 45 cos^{2} (x) - 7

- 16 - 36 cos^{2} (x) + 48 cos (x) + 9 - 9 cos^{2} (x)

对同类项分组

= - 36 cos^{2} (x) + 48 cos (x) - 9 cos^{2} (x) - 16 + 9

同类项相加:

- 36 cos^{2} (x) - 9 cos^{2} (x) = - 45 cos^{2} (x)

= - 45 cos^{2} (x) + 48 cos (x) - 16 + 9

数字相加/相减:

- 16 + 9 = - 7

= 48 cos (x) - 45 cos^{2} (x) - 7

= 48 cos (x) - 45 cos^{2} (x) - 7

= 48 cos (x) - 45 cos^{2} (x) - 7

- 7 - 45 cos^{2} (x) + 48 cos (x) = 0

用替代法求解

- 7 - 45 cos^{2} (x) + 48 cos (x) = 0

令

: cos (x) = u

- 7 - 45 u^{2} + 48 u = 0

- 7 - 45 u^{2} + 48 u = 0 : u = \frac{8 - 29}{15}, u = \frac{8 + 29}{15}

- 7 - 45 u^{2} + 48 u = 0

改写成标准形式

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

- 45 u^{2} + 48 u - 7 = 0

使用求根公式求解

- 45 u^{2} + 48 u - 7 = 0

二次方程求根公式:

若

a = - 45, b = 48, c = - 7

u_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 45 ) ( - 7 )}{2 ( - 45 )}

u_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 45 ) ( - 7 )}{2 ( - 45 )}

4 8^{2} - 4 (- 45) (- 7) = 629

4 8^{2} - 4 (- 45) (- 7)

使用法则

- (- a) = a

= 4 8^{2} - 4 \cdot 45 \cdot 7

数字相乘:

4 \cdot 45 \cdot 7 = 1260

= 4 8^{2} - 1260

4 8^{2} = 2304

= 2304 - 1260

数字相减:

2304 - 1260 = 1044

= 1044

1044

质因数分解:

2^{2} \cdot 3^{2} \cdot 29

1044

1044

除以

21044 = 522 \cdot 2

= 2 \cdot 522

522

除以

2522 = 261 \cdot 2

= 2 \cdot 2 \cdot 261

261

除以

3261 = 87 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 87

87

除以

387 = 29 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 29

2, 3, 29

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 29

= 2^{2} \cdot 3^{2} \cdot 29

= 2^{2} \cdot 3^{2} \cdot 29

使用根式运算法则:

n ab = n a n b

= 29 2^{2} 3^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 229 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 329

整理后得

= 629

u_{1, 2} = \frac{- 48 \pm 6 29}{2 ( - 45 )}

将解分隔开

u_{1} = \frac{- 48 + 6 29}{2 ( - 45 )}, u_{2} = \frac{- 48 - 6 29}{2 ( - 45 )}

u = \frac{- 48 + 6 29}{2 ( - 45 )} : \frac{8 - 29}{15}

\frac{- 48 + 6 29}{2 ( - 45 )}

去除括号:

(- a) = - a

= \frac{- 48 + 6 29}{- 2 \cdot 45}

数字相乘:

2 \cdot 45 = 90

= \frac{- 48 + 6 29}{- 90}

使用分式法则:

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

- 48 + 629 = - (48 - 629)

= \frac{48 - 6 29}{90}

分解

48 - 629 : 6 (8 - 29)

48 - 629

改写为

= 6 \cdot 8 - 629

因式分解出通项

6

= 6 (8 - 29)

= \frac{6 ( 8 - 29 )}{90}

约分:

6

= \frac{8 - 29}{15}

u = \frac{- 48 - 6 29}{2 ( - 45 )} : \frac{8 + 29}{15}

\frac{- 48 - 6 29}{2 ( - 45 )}

去除括号:

(- a) = - a

= \frac{- 48 - 6 29}{- 2 \cdot 45}

数字相乘:

2 \cdot 45 = 90

= \frac{- 48 - 6 29}{- 90}

使用分式法则:

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

- 48 - 629 = - (48 + 629)

= \frac{48 + 6 29}{90}

分解

48 + 629 : 6 (8 + 29)

48 + 629

改写为

= 6 \cdot 8 + 629

因式分解出通项

6

= 6 (8 + 29)

= \frac{6 ( 8 + 29 )}{90}

约分:

6

= \frac{8 + 29}{15}

二次方程组的解是:

u = \frac{8 - 29}{15}, u = \frac{8 + 29}{15}

u = cos (x)

代回

cos (x) = \frac{8 - 29}{15}, cos (x) = \frac{8 + 29}{15}

cos (x) = \frac{8 - 29}{15}, cos (x) = \frac{8 + 29}{15}

cos (x) = \frac{8 - 29}{15} : x = arccos (\frac{8 - 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 - 29}{15}) + 2 πn

cos (x) = \frac{8 - 29}{15}

使用反三角函数性质

cos (x) = \frac{8 - 29}{15}

cos (x) = \frac{8 - 29}{15}

的通解

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

x = arccos (\frac{8 - 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 - 29}{15}) + 2 πn

x = arccos (\frac{8 - 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 - 29}{15}) + 2 πn

cos (x) = \frac{8 + 29}{15} : x = arccos (\frac{8 + 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 + 29}{15}) + 2 πn

cos (x) = \frac{8 + 29}{15}

使用反三角函数性质

cos (x) = \frac{8 + 29}{15}

cos (x) = \frac{8 + 29}{15}

的通解

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

x = arccos (\frac{8 + 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 + 29}{15}) + 2 πn

x = arccos (\frac{8 + 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 + 29}{15}) + 2 πn

合并所有解

x = arccos (\frac{8 - 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 - 29}{15}) + 2 πn, x = arccos (\frac{8 + 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 + 29}{15}) + 2 πn

将解代入原方程进行验证

将它们代入

3 sin (x) + 6 cos (x) = 4

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{8 - 29}{15}) + 2 πn

的解:

真

arccos (\frac{8 - 29}{15}) + 2 πn

代入

n = 1

arccos (\frac{8 - 29}{15}) + 2 π 1

对于

3 sin (x) + 6 cos (x) = 4

代入

x = arccos (\frac{8 - 29}{15}) + 2 π 1

3 sin (arccos (\frac{8 - 29}{15}) + 2 π 1) + 6 cos (arccos (\frac{8 - 29}{15}) + 2 π 1) = 4

整理后得

4 = 4

\Rightarrow 真

检验

2 π - arccos (\frac{8 - 29}{15}) + 2 πn

的解:

假

2 π - arccos (\frac{8 - 29}{15}) + 2 πn

代入

n = 1

2 π - arccos (\frac{8 - 29}{15}) + 2 π 1

对于

3 sin (x) + 6 cos (x) = 4

代入

x = 2 π - arccos (\frac{8 - 29}{15}) + 2 π 1

3 sin (2 π - arccos (\frac{8 - 29}{15}) + 2 π 1) + 6 cos (2 π - arccos (\frac{8 - 29}{15}) + 2 π 1) = 4

整理后得

- 1.90813 \dots = 4

\Rightarrow 假

检验

arccos (\frac{8 + 29}{15}) + 2 πn

的解:

假

arccos (\frac{8 + 29}{15}) + 2 πn

代入

n = 1

arccos (\frac{8 + 29}{15}) + 2 π 1

对于

3 sin (x) + 6 cos (x) = 4

代入

x = arccos (\frac{8 + 29}{15}) + 2 π 1

3 sin (arccos (\frac{8 + 29}{15}) + 2 π 1) + 6 cos (arccos (\frac{8 + 29}{15}) + 2 π 1) = 4

整理后得

6.70813 \dots = 4

\Rightarrow 假

检验

2 π - arccos (\frac{8 + 29}{15}) + 2 πn

的解:

真

2 π - arccos (\frac{8 + 29}{15}) + 2 πn

代入

n = 1

2 π - arccos (\frac{8 + 29}{15}) + 2 π 1

对于

3 sin (x) + 6 cos (x) = 4

代入

x = 2 π - arccos (\frac{8 + 29}{15}) + 2 π 1

3 sin (2 π - arccos (\frac{8 + 29}{15}) + 2 π 1) + 6 cos (2 π - arccos (\frac{8 + 29}{15}) + 2 π 1) = 4

整理后得

4 = 4

\Rightarrow 真

x = arccos (\frac{8 - 29}{15}) + 2 πn, x = 2 π - arccos (\frac{8 + 29}{15}) + 2 πn

以小数形式表示解

x = 1.39557 \dots + 2 πn, x = 2 π - 0.46828 \dots + 2 πn

作图

流行的例子

3 tan (2 x + 1) = 2

tan(2x)-2=3tan(x)

tan (2 x) - 2 = 3 tan (x)

sin (x - \frac{π}{6}) = 0

arccot(x-2)=arccot(x-1)+arccot(x)

arccot (x - 2) = arccot (x - 1) + arccot (x)

arccos (2 x) = π