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受欢迎的三角函数 >

2sin(x)-4cos(x)=3,0<= x<= 2pi

解答

2 sin (x) - 4 cos (x) = 3, 0 \leq x \leq 2 π

解答

x = - 2.76975 \dots + 2 π, x = 1.84246 \dots

+1

度数

x = 201.30453 \dots^{\circ}, x = 105.56536 \dots^{\circ}

求解步骤

2 sin (x) - 4 cos (x) = 3, 0 \leq x \leq 2 π

两边加上

4 cos (x)

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

两边进行平方

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

两边减去

(3 + 4 cos (x))^{2}

4 sin^{2} (x) - 9 - 24 cos (x) - 16 cos^{2} (x) = 0

使用三角恒等式改写

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 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)

= - 9 - 16 cos^{2} (x) - 24 cos (x) + 4 (1 - cos^{2} (x))

化简

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 (1 - cos^{2} (x)) : - 20 cos^{2} (x) - 24 cos (x) - 5

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 (1 - cos^{2} (x))

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= - 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x)

化简

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x) : - 20 cos^{2} (x) - 24 cos (x) - 5

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x)

对同类项分组

= - 16 cos^{2} (x) - 24 cos (x) - 4 cos^{2} (x) - 9 + 4

同类项相加:

- 16 cos^{2} (x) - 4 cos^{2} (x) = - 20 cos^{2} (x)

= - 20 cos^{2} (x) - 24 cos (x) - 9 + 4

数字相加/相减:

- 9 + 4 = - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

- 5 - 20 cos^{2} (x) - 24 cos (x) = 0

用替代法求解

- 5 - 20 cos^{2} (x) - 24 cos (x) = 0

令

: cos (x) = u

- 5 - 20 u^{2} - 24 u = 0

- 5 - 20 u^{2} - 24 u = 0 : u = - \frac{6 + 11}{10}, u = - \frac{6 - 11}{10}

- 5 - 20 u^{2} - 24 u = 0

改写成标准形式

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

- 20 u^{2} - 24 u - 5 = 0

使用求根公式求解

- 20 u^{2} - 24 u - 5 = 0

二次方程求根公式:

若

a = - 20, b = - 24, c = - 5

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 ( - 20 ) ( - 5 )}{2 ( - 20 )}

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 ( - 20 ) ( - 5 )}{2 ( - 20 )}

(- 24)^{2} - 4 (- 20) (- 5) = 411

(- 24)^{2} - 4 (- 20) (- 5)

使用法则

- (- a) = a

= (- 24)^{2} - 4 \cdot 20 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 24)^{2} = 2 4^{2}

= 2 4^{2} - 4 \cdot 20 \cdot 5

数字相乘:

4 \cdot 20 \cdot 5 = 400

= 2 4^{2} - 400

2 4^{2} = 576

= 576 - 400

数字相减:

576 - 400 = 176

= 176

176

质因数分解:

2^{4} \cdot 11

176

176

除以

2176 = 88 \cdot 2

= 2 \cdot 88

88

除以

288 = 44 \cdot 2

= 2 \cdot 2 \cdot 44

44

除以

244 = 22 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 22

22

除以

222 = 11 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 11

2, 11

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 11

= 2^{4} \cdot 11

= 2^{4} \cdot 11

使用根式运算法则:

n ab = n a n b

= 11 2^{4}

使用根式运算法则:

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

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

= 2^{2} 11

整理后得

= 411

u_{1, 2} = \frac{- ( - 24 ) \pm 4 11}{2 ( - 20 )}

将解分隔开

u_{1} = \frac{- ( - 24 ) + 4 11}{2 ( - 20 )}, u_{2} = \frac{- ( - 24 ) - 4 11}{2 ( - 20 )}

u = \frac{- ( - 24 ) + 4 11}{2 ( - 20 )} : - \frac{6 + 11}{10}

\frac{- ( - 24 ) + 4 11}{2 ( - 20 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{24 + 4 11}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{24 + 4 11}{- 40}

使用分式法则:

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

= - \frac{24 + 4 11}{40}

消掉

\frac{24 + 4 11}{40} : \frac{6 + 11}{10}

\frac{24 + 4 11}{40}

分解

24 + 411 : 4 (6 + 11)

24 + 411

改写为

= 4 \cdot 6 + 411

因式分解出通项

4

= 4 (6 + 11)

= \frac{4 ( 6 + 11 )}{40}

约分:

4

= \frac{6 + 11}{10}

= - \frac{6 + 11}{10}

u = \frac{- ( - 24 ) - 4 11}{2 ( - 20 )} : - \frac{6 - 11}{10}

\frac{- ( - 24 ) - 4 11}{2 ( - 20 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{24 - 4 11}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{24 - 4 11}{- 40}

使用分式法则:

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

= - \frac{24 - 4 11}{40}

消掉

\frac{24 - 4 11}{40} : \frac{6 - 11}{10}

\frac{24 - 4 11}{40}

分解

24 - 411 : 4 (6 - 11)

24 - 411

改写为

= 4 \cdot 6 - 411

因式分解出通项

4

= 4 (6 - 11)

= \frac{4 ( 6 - 11 )}{40}

约分:

4

= \frac{6 - 11}{10}

= - \frac{6 - 11}{10}

二次方程组的解是:

u = - \frac{6 + 11}{10}, u = - \frac{6 - 11}{10}

u = cos (x)

代回

cos (x) = - \frac{6 + 11}{10}, cos (x) = - \frac{6 - 11}{10}

cos (x) = - \frac{6 + 11}{10}, cos (x) = - \frac{6 - 11}{10}

cos (x) = - \frac{6 + 11}{10}, 0 \leq x \leq 2 π : x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π

cos (x) = - \frac{6 + 11}{10}, 0 \leq x \leq 2 π

使用反三角函数性质

cos (x) = - \frac{6 + 11}{10}

cos (x) = - \frac{6 + 11}{10}

的通解

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

x = arccos (- \frac{6 + 11}{10}) + 2 πn, x = - arccos (- \frac{6 + 11}{10}) + 2 πn

x = arccos (- \frac{6 + 11}{10}) + 2 πn, x = - arccos (- \frac{6 + 11}{10}) + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π

cos (x) = - \frac{6 - 11}{10}, 0 \leq x \leq 2 π : x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

cos (x) = - \frac{6 - 11}{10}, 0 \leq x \leq 2 π

使用反三角函数性质

cos (x) = - \frac{6 - 11}{10}

cos (x) = - \frac{6 - 11}{10}

的通解

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

x = arccos (- \frac{6 - 11}{10}) + 2 πn, x = - arccos (- \frac{6 - 11}{10}) + 2 πn

x = arccos (- \frac{6 - 11}{10}) + 2 πn, x = - arccos (- \frac{6 - 11}{10}) + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

合并所有解

x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π, x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{6 + 11}{10})

的解:

假

arccos (- \frac{6 + 11}{10})

代入

n = 1

arccos (- \frac{6 + 11}{10})

对于

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

代入

x = arccos (- \frac{6 + 11}{10})

2 sin (arccos (- \frac{6 + 11}{10})) - 4 cos (arccos (- \frac{6 + 11}{10})) = 3

整理后得

4.45329 \dots = 3

\Rightarrow 假

检验

- arccos (- \frac{6 + 11}{10}) + 2 π

的解:

真

- arccos (- \frac{6 + 11}{10}) + 2 π

代入

n = 1

- arccos (- \frac{6 + 11}{10}) + 2 π

对于

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

代入

x = - arccos (- \frac{6 + 11}{10}) + 2 π

2 sin (- arccos (- \frac{6 + 11}{10}) + 2 π) - 4 cos (- arccos (- \frac{6 + 11}{10}) + 2 π) = 3

整理后得

3 = 3

\Rightarrow 真

检验

arccos (- \frac{6 - 11}{10})

的解:

真

arccos (- \frac{6 - 11}{10})

代入

n = 1

arccos (- \frac{6 - 11}{10})

对于

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

代入

x = arccos (- \frac{6 - 11}{10})

2 sin (arccos (- \frac{6 - 11}{10})) - 4 cos (arccos (- \frac{6 - 11}{10})) = 3

整理后得

3 = 3

\Rightarrow 真

检验

- arccos (- \frac{6 - 11}{10}) + 2 π

的解:

假

- arccos (- \frac{6 - 11}{10}) + 2 π

代入

n = 1

- arccos (- \frac{6 - 11}{10}) + 2 π

对于

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

代入

x = - arccos (- \frac{6 - 11}{10}) + 2 π

2 sin (- arccos (- \frac{6 - 11}{10}) + 2 π) - 4 cos (- arccos (- \frac{6 - 11}{10}) + 2 π) = 3

整理后得

- 0.85329 \dots = 3

\Rightarrow 假

x = - arccos (- \frac{6 + 11}{10}) + 2 π, x = arccos (- \frac{6 - 11}{10})

以小数形式表示解

x = - 2.76975 \dots + 2 π, x = 1.84246 \dots

作图

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