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受欢迎的三角函数 >

15sin(x)+6cos(x)-3=0

解答

15 sin (x) + 6 cos (x) - 3 = 0

解答

x = π - 0.56728 \dots + 2 πn, x = - 0.19372 \dots + 2 πn

+1

度数

x = 147.49691 \dots^{\circ} + 36 0^{\circ} n, x = - 11.09973 \dots^{\circ} + 36 0^{\circ} n

求解步骤

15 sin (x) + 6 cos (x) - 3 = 0

两边减去

6 cos (x)

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

两边进行平方

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

两边减去

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

(15 sin (x) - 3)^{2} - 36 cos^{2} (x) = 0

使用三角恒等式改写

(- 3 + 15 sin (x))^{2} - 36 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= (- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x))

化简

(- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x)) : 261 sin^{2} (x) - 90 sin (x) - 27

(- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x))

(- 3 + 15 sin (x))^{2} : 9 - 90 sin (x) + 225 sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - 3, b = 15 sin (x)

= (- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2}

化简

(- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2} : 9 - 90 sin (x) + 225 sin^{2} (x)

(- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2}

去除括号:

(- a) = - a

= (- 3)^{2} - 2 \cdot 3 \cdot 15 sin (x) + (15 sin (x))^{2}

(- 3)^{2} = 9

(- 3)^{2}

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2}

3^{2} = 9

= 9

2 \cdot 3 \cdot 15 sin (x) = 90 sin (x)

2 \cdot 3 \cdot 15 sin (x)

数字相乘:

2 \cdot 3 \cdot 15 = 90

= 90 sin (x)

(15 sin (x))^{2} = 225 sin^{2} (x)

(15 sin (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 1 5^{2} sin^{2} (x)

1 5^{2} = 225

= 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x) - 36 (1 - sin^{2} (x))

乘开

- 36 (1 - sin^{2} (x)) : - 36 + 36 sin^{2} (x)

- 36 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 36, b = 1, c = sin^{2} (x)

= - 36 \cdot 1 - (- 36) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 36 \cdot 1 + 36 sin^{2} (x)

数字相乘:

36 \cdot 1 = 36

= - 36 + 36 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x)

化简

9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x) : 261 sin^{2} (x) - 90 sin (x) - 27

9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x)

对同类项分组

= - 90 sin (x) + 225 sin^{2} (x) + 36 sin^{2} (x) + 9 - 36

同类项相加:

225 sin^{2} (x) + 36 sin^{2} (x) = 261 sin^{2} (x)

= - 90 sin (x) + 261 sin^{2} (x) + 9 - 36

数字相加/相减:

9 - 36 = - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

- 27 + 261 sin^{2} (x) - 90 sin (x) = 0

用替代法求解

- 27 + 261 sin^{2} (x) - 90 sin (x) = 0

令

: sin (x) = u

- 27 + 261 u^{2} - 90 u = 0

- 27 + 261 u^{2} - 90 u = 0 : u = \frac{5 + 4 7}{29}, u = \frac{5 - 4 7}{29}

- 27 + 261 u^{2} - 90 u = 0

改写成标准形式

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

261 u^{2} - 90 u - 27 = 0

使用求根公式求解

261 u^{2} - 90 u - 27 = 0

二次方程求根公式:

若

a = 261, b = - 90, c = - 27

u_{1, 2} = \frac{- ( - 90 ) \pm ( - 90 ) ^{2} - 4 \cdot 261 ( - 27 )}{2 \cdot 261}

u_{1, 2} = \frac{- ( - 90 ) \pm ( - 90 ) ^{2} - 4 \cdot 261 ( - 27 )}{2 \cdot 261}

(- 90)^{2} - 4 \cdot 261 (- 27) = 727

(- 90)^{2} - 4 \cdot 261 (- 27)

使用法则

- (- a) = a

= (- 90)^{2} + 4 \cdot 261 \cdot 27

使用指数法则:

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

若

n

是偶数

(- 90)^{2} = 9 0^{2}

= 9 0^{2} + 4 \cdot 261 \cdot 27

数字相乘:

4 \cdot 261 \cdot 27 = 28188

= 9 0^{2} + 28188

9 0^{2} = 8100

= 8100 + 28188

数字相加:

8100 + 28188 = 36288

= 36288

36288

质因数分解:

2^{6} \cdot 3^{4} \cdot 7

36288

36288

除以

236288 = 18144 \cdot 2

= 2 \cdot 18144

18144

除以

218144 = 9072 \cdot 2

= 2 \cdot 2 \cdot 9072

9072

除以

29072 = 4536 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4536

4536

除以

24536 = 2268 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2268

2268

除以

22268 = 1134 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 1134

1134

除以

21134 = 567 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 567

567

除以

3567 = 189 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 189

189

除以

3189 = 63 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 63

63

除以

363 = 21 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 21

21

除以

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 7

2, 3, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 7

= 2^{6} \cdot 3^{4} \cdot 7

= 2^{6} \cdot 3^{4} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{6} 3^{4}

使用根式运算法则:

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

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 7 3^{4}

使用根式运算法则:

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

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 2^{3} \cdot 3^{2} 7

整理后得

= 727

u_{1, 2} = \frac{- ( - 90 ) \pm 72 7}{2 \cdot 261}

将解分隔开

u_{1} = \frac{- ( - 90 ) + 72 7}{2 \cdot 261}, u_{2} = \frac{- ( - 90 ) - 72 7}{2 \cdot 261}

u = \frac{- ( - 90 ) + 72 7}{2 \cdot 261} : \frac{5 + 4 7}{29}

\frac{- ( - 90 ) + 72 7}{2 \cdot 261}

使用法则

- (- a) = a

= \frac{90 + 72 7}{2 \cdot 261}

数字相乘:

2 \cdot 261 = 522

= \frac{90 + 72 7}{522}

分解

90 + 727 : 18 (5 + 47)

90 + 727

改写为

= 18 \cdot 5 + 18 \cdot 47

因式分解出通项

18

= 18 (5 + 47)

= \frac{18 ( 5 + 4 7 )}{522}

约分:

18

= \frac{5 + 4 7}{29}

u = \frac{- ( - 90 ) - 72 7}{2 \cdot 261} : \frac{5 - 4 7}{29}

\frac{- ( - 90 ) - 72 7}{2 \cdot 261}

使用法则

- (- a) = a

= \frac{90 - 72 7}{2 \cdot 261}

数字相乘:

2 \cdot 261 = 522

= \frac{90 - 72 7}{522}

分解

90 - 727 : 18 (5 - 47)

90 - 727

改写为

= 18 \cdot 5 - 18 \cdot 47

因式分解出通项

18

= 18 (5 - 47)

= \frac{18 ( 5 - 4 7 )}{522}

约分:

18

= \frac{5 - 4 7}{29}

二次方程组的解是:

u = \frac{5 + 4 7}{29}, u = \frac{5 - 4 7}{29}

u = sin (x)

代回

sin (x) = \frac{5 + 4 7}{29}, sin (x) = \frac{5 - 4 7}{29}

sin (x) = \frac{5 + 4 7}{29}, sin (x) = \frac{5 - 4 7}{29}

sin (x) = \frac{5 + 4 7}{29} : x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

sin (x) = \frac{5 + 4 7}{29}

使用反三角函数性质

sin (x) = \frac{5 + 4 7}{29}

sin (x) = \frac{5 + 4 7}{29}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

sin (x) = \frac{5 - 4 7}{29} : x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

sin (x) = \frac{5 - 4 7}{29}

使用反三角函数性质

sin (x) = \frac{5 - 4 7}{29}

sin (x) = \frac{5 - 4 7}{29}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

合并所有解

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

将解代入原方程进行验证

将它们代入

15 sin (x) + 6 cos (x) - 3 = 0

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{5 + 4 7}{29}) + 2 πn

的解:

假

arcsin (\frac{5 + 4 7}{29}) + 2 πn

代入

n = 1

arcsin (\frac{5 + 4 7}{29}) + 2 π 1

对于

15 sin (x) + 6 cos (x) - 3 = 0

代入

x = arcsin (\frac{5 + 4 7}{29}) + 2 π 1

15 sin (arcsin (\frac{5 + 4 7}{29}) + 2 π 1) + 6 cos (arcsin (\frac{5 + 4 7}{29}) + 2 π 1) - 3 = 0

整理后得

10.12035 \dots = 0

\Rightarrow 假

检验

π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

的解:

真

π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

代入

n = 1

π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1

对于

15 sin (x) + 6 cos (x) - 3 = 0

代入

x = π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1

15 sin (π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1) + 6 cos (π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1) - 3 = 0

整理后得

0 = 0

\Rightarrow 真

检验

arcsin (\frac{5 - 4 7}{29}) + 2 πn

的解:

真

arcsin (\frac{5 - 4 7}{29}) + 2 πn

代入

n = 1

arcsin (\frac{5 - 4 7}{29}) + 2 π 1

对于

15 sin (x) + 6 cos (x) - 3 = 0

代入

x = arcsin (\frac{5 - 4 7}{29}) + 2 π 1

15 sin (arcsin (\frac{5 - 4 7}{29}) + 2 π 1) + 6 cos (arcsin (\frac{5 - 4 7}{29}) + 2 π 1) - 3 = 0

整理后得

0 = 0

\Rightarrow 真

检验

π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

的解:

假

π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

代入

n = 1

π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1

对于

15 sin (x) + 6 cos (x) - 3 = 0

代入

x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1

15 sin (π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1) + 6 cos (π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1) - 3 = 0

整理后得

- 11.77552 \dots = 0

\Rightarrow 假

x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = arcsin (\frac{5 - 4 7}{29}) + 2 πn

以小数形式表示解

x = π - 0.56728 \dots + 2 πn, x = - 0.19372 \dots + 2 πn

作图

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