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受欢迎的三角函数 >

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

解答

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

解答

x = 1.80278 \dots + 2 πn, x = 2 π - 0.51577 \dots + 2 πn

+1

度数

x = 103.29171 \dots^{\circ} + 36 0^{\circ} n, x = 330.44807 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

4 cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

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

化简

- 4 + 16 cos (x) - 16 cos^{2} (x) + 9 (1 - cos^{2} (x)) : 16 cos (x) - 25 cos^{2} (x) + 5

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

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

化简

- 4 + 16 cos (x) - 16 cos^{2} (x) + 9 - 9 cos^{2} (x) : 16 cos (x) - 25 cos^{2} (x) + 5

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

对同类项分组

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

同类项相加:

- 16 cos^{2} (x) - 9 cos^{2} (x) = - 25 cos^{2} (x)

= 16 cos (x) - 25 cos^{2} (x) - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= 16 cos (x) - 25 cos^{2} (x) + 5

= 16 cos (x) - 25 cos^{2} (x) + 5

= 16 cos (x) - 25 cos^{2} (x) + 5

5 + 16 cos (x) - 25 cos^{2} (x) = 0

用替代法求解

5 + 16 cos (x) - 25 cos^{2} (x) = 0

令

: cos (x) = u

5 + 16 u - 25 u^{2} = 0

5 + 16 u - 25 u^{2} = 0 : u = - \frac{- 8 + 3 21}{25}, u = \frac{8 + 3 21}{25}

5 + 16 u - 25 u^{2} = 0

改写成标准形式

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

- 25 u^{2} + 16 u + 5 = 0

使用求根公式求解

- 25 u^{2} + 16 u + 5 = 0

二次方程求根公式:

若

a = - 25, b = 16, c = 5

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 25 ) \cdot 5}{2 ( - 25 )}

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 25 ) \cdot 5}{2 ( - 25 )}

1 6^{2} - 4 (- 25) \cdot 5 = 621

1 6^{2} - 4 (- 25) \cdot 5

使用法则

- (- a) = a

= 1 6^{2} + 4 \cdot 25 \cdot 5

数字相乘:

4 \cdot 25 \cdot 5 = 500

= 1 6^{2} + 500

1 6^{2} = 256

= 256 + 500

数字相加:

256 + 500 = 756

= 756

756

质因数分解:

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

756

756

除以

2756 = 378 \cdot 2

= 2 \cdot 378

378

除以

2378 = 189 \cdot 2

= 2 \cdot 2 \cdot 189

189

除以

3189 = 63 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 63

63

除以

363 = 21 \cdot 3

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

21

除以

321 = 7 \cdot 3

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

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

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

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

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

使用根式运算法则:

n ab = n a n b

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

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 3 \cdot 7

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 3 \cdot 7

整理后得

= 621

u_{1, 2} = \frac{- 16 \pm 6 21}{2 ( - 25 )}

将解分隔开

u_{1} = \frac{- 16 + 6 21}{2 ( - 25 )}, u_{2} = \frac{- 16 - 6 21}{2 ( - 25 )}

u = \frac{- 16 + 6 21}{2 ( - 25 )} : - \frac{- 8 + 3 21}{25}

\frac{- 16 + 6 21}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 16 + 6 21}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{- 16 + 6 21}{- 50}

使用分式法则:

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

= - \frac{- 16 + 6 21}{50}

消掉

\frac{- 16 + 6 21}{50} : \frac{3 21 - 8}{25}

\frac{- 16 + 6 21}{50}

分解

- 16 + 621 : 2 (- 8 + 321)

- 16 + 621

改写为

= - 2 \cdot 8 + 2 \cdot 321

因式分解出通项

2

= 2 (- 8 + 321)

= \frac{2 ( - 8 + 3 21 )}{50}

约分:

2

= \frac{- 8 + 3 21}{25}

= - \frac{3 21 - 8}{25}

= - \frac{- 8 + 3 21}{25}

u = \frac{- 16 - 6 21}{2 ( - 25 )} : \frac{8 + 3 21}{25}

\frac{- 16 - 6 21}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 16 - 6 21}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{- 16 - 6 21}{- 50}

使用分式法则:

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

- 16 - 621 = - (16 + 621)

= \frac{16 + 6 21}{50}

分解

16 + 621 : 2 (8 + 321)

16 + 621

改写为

= 2 \cdot 8 + 2 \cdot 321

因式分解出通项

2

= 2 (8 + 321)

= \frac{2 ( 8 + 3 21 )}{50}

约分:

2

= \frac{8 + 3 21}{25}

二次方程组的解是:

u = - \frac{- 8 + 3 21}{25}, u = \frac{8 + 3 21}{25}

u = cos (x)

代回

cos (x) = - \frac{- 8 + 3 21}{25}, cos (x) = \frac{8 + 3 21}{25}

cos (x) = - \frac{- 8 + 3 21}{25}, cos (x) = \frac{8 + 3 21}{25}

cos (x) = - \frac{- 8 + 3 21}{25} : x = arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = - arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

cos (x) = - \frac{- 8 + 3 21}{25}

使用反三角函数性质

cos (x) = - \frac{- 8 + 3 21}{25}

cos (x) = - \frac{- 8 + 3 21}{25}

的通解

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

x = arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = - arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

x = arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = - arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

cos (x) = \frac{8 + 3 21}{25} : x = arccos (\frac{8 + 3 21}{25}) + 2 πn, x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

cos (x) = \frac{8 + 3 21}{25}

使用反三角函数性质

cos (x) = \frac{8 + 3 21}{25}

cos (x) = \frac{8 + 3 21}{25}

的通解

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

x = arccos (\frac{8 + 3 21}{25}) + 2 πn, x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

x = arccos (\frac{8 + 3 21}{25}) + 2 πn, x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

合并所有解

x = arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = - arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = arccos (\frac{8 + 3 21}{25}) + 2 πn, x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

的解:

真

arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

代入

n = 1

arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1

对于

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

代入

x = arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1

3 sin (arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1) + 4 cos (arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

- arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

的解:

假

- arccos (- \frac{- 8 + 3 21}{25}) + 2 πn

代入

n = 1

- arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1

3 sin (- arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1) + 4 cos (- arccos (- \frac{- 8 + 3 21}{25}) + 2 π 1) = 2

整理后得

- 3.83927 \dots = 2

\Rightarrow 假

检验

arccos (\frac{8 + 3 21}{25}) + 2 πn

的解:

假

arccos (\frac{8 + 3 21}{25}) + 2 πn

代入

n = 1

arccos (\frac{8 + 3 21}{25}) + 2 π 1

对于

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

代入

x = arccos (\frac{8 + 3 21}{25}) + 2 π 1

3 sin (arccos (\frac{8 + 3 21}{25}) + 2 π 1) + 4 cos (arccos (\frac{8 + 3 21}{25}) + 2 π 1) = 2

整理后得

4.95927 \dots = 2

\Rightarrow 假

检验

2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

的解:

真

2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

代入

n = 1

2 π - arccos (\frac{8 + 3 21}{25}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 π 1

3 sin (2 π - arccos (\frac{8 + 3 21}{25}) + 2 π 1) + 4 cos (2 π - arccos (\frac{8 + 3 21}{25}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

x = arccos (- \frac{- 8 + 3 21}{25}) + 2 πn, x = 2 π - arccos (\frac{8 + 3 21}{25}) + 2 πn

以小数形式表示解

x = 1.80278 \dots + 2 πn, x = 2 π - 0.51577 \dots + 2 πn

作图

流行的例子

cot^2(x)-6cot(x)+5=0,0<= x<= 2pi

cot^{2} (x) - 6 cot (x) + 5 = 0, 0 \leq x \leq 2 π

9tan(θ)+13=2tan(θ)+6

9 tan (θ) + 13 = 2 tan (θ) + 6

2 = 3 sin (∣ 5 x ∣) + 1

sin (θ) = - \frac{1}{3}

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