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受欢迎的三角函数 >

7sin(x)=cos(x)-3

解答

7 sin (x) = cos (x) - 3

解答

x = - 2.56154 \dots + 2 πn, x = 2 π - 0.29625 \dots + 2 πn

+1

度数

x = - 146.76580 \dots^{\circ} + 36 0^{\circ} n, x = 343.02601 \dots^{\circ} + 36 0^{\circ} n

求解步骤

7 sin (x) = cos (x) - 3

两边进行平方

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

两边减去

(cos (x) - 3)^{2}

49 sin^{2} (x) - cos^{2} (x) + 6 cos (x) - 9 = 0

使用三角恒等式改写

- 9 - cos^{2} (x) + 49 sin^{2} (x) + 6 cos (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 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 6 cos (x)

化简

- 9 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 6 cos (x) : 6 cos (x) - 50 cos^{2} (x) + 40

- 9 - cos^{2} (x) + 49 (1 - cos^{2} (x)) + 6 cos (x)

乘开

49 (1 - cos^{2} (x)) : 49 - 49 cos^{2} (x)

49 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 49, b = 1, c = cos^{2} (x)

= 49 \cdot 1 - 49 cos^{2} (x)

数字相乘:

49 \cdot 1 = 49

= 49 - 49 cos^{2} (x)

= - 9 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 6 cos (x)

化简

- 9 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 6 cos (x) : 6 cos (x) - 50 cos^{2} (x) + 40

- 9 - cos^{2} (x) + 49 - 49 cos^{2} (x) + 6 cos (x)

对同类项分组

= - cos^{2} (x) - 49 cos^{2} (x) + 6 cos (x) - 9 + 49

同类项相加:

- cos^{2} (x) - 49 cos^{2} (x) = - 50 cos^{2} (x)

= - 50 cos^{2} (x) + 6 cos (x) - 9 + 49

数字相加/相减:

- 9 + 49 = 40

= 6 cos (x) - 50 cos^{2} (x) + 40

= 6 cos (x) - 50 cos^{2} (x) + 40

= 6 cos (x) - 50 cos^{2} (x) + 40

40 - 50 cos^{2} (x) + 6 cos (x) = 0

用替代法求解

40 - 50 cos^{2} (x) + 6 cos (x) = 0

令

: cos (x) = u

40 - 50 u^{2} + 6 u = 0

40 - 50 u^{2} + 6 u = 0 : u = - \frac{- 3 + 7 41}{50}, u = \frac{3 + 7 41}{50}

40 - 50 u^{2} + 6 u = 0

改写成标准形式

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

- 50 u^{2} + 6 u + 40 = 0

使用求根公式求解

- 50 u^{2} + 6 u + 40 = 0

二次方程求根公式:

若

a = - 50, b = 6, c = 40

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 50 ) \cdot 40}{2 ( - 50 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 50 ) \cdot 40}{2 ( - 50 )}

6^{2} - 4 (- 50) \cdot 40 = 1441

6^{2} - 4 (- 50) \cdot 40

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 50 \cdot 40

数字相乘:

4 \cdot 50 \cdot 40 = 8000

= 6^{2} + 8000

6^{2} = 36

= 36 + 8000

数字相加:

36 + 8000 = 8036

= 8036

8036

质因数分解:

2^{2} \cdot 7^{2} \cdot 41

8036

8036

除以

28036 = 4018 \cdot 2

= 2 \cdot 4018

4018

除以

24018 = 2009 \cdot 2

= 2 \cdot 2 \cdot 2009

2009

除以

72009 = 287 \cdot 7

= 2 \cdot 2 \cdot 7 \cdot 287

287

除以

7287 = 41 \cdot 7

= 2 \cdot 2 \cdot 7 \cdot 7 \cdot 41

2, 7, 41

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7 \cdot 7 \cdot 41

= 2^{2} \cdot 7^{2} \cdot 41

= 2^{2} \cdot 7^{2} \cdot 41

使用根式运算法则:

n ab = n a n b

= 41 2^{2} 7^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 241 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 2 \cdot 741

整理后得

= 1441

u_{1, 2} = \frac{- 6 \pm 14 41}{2 ( - 50 )}

将解分隔开

u_{1} = \frac{- 6 + 14 41}{2 ( - 50 )}, u_{2} = \frac{- 6 - 14 41}{2 ( - 50 )}

u = \frac{- 6 + 14 41}{2 ( - 50 )} : - \frac{- 3 + 7 41}{50}

\frac{- 6 + 14 41}{2 ( - 50 )}

去除括号:

(- a) = - a

= \frac{- 6 + 14 41}{- 2 \cdot 50}

数字相乘:

2 \cdot 50 = 100

= \frac{- 6 + 14 41}{- 100}

使用分式法则:

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

= - \frac{- 6 + 14 41}{100}

消掉

\frac{- 6 + 14 41}{100} : \frac{7 41 - 3}{50}

\frac{- 6 + 14 41}{100}

分解

- 6 + 1441 : 2 (- 3 + 741)

- 6 + 1441

改写为

= - 2 \cdot 3 + 2 \cdot 741

因式分解出通项

2

= 2 (- 3 + 741)

= \frac{2 ( - 3 + 7 41 )}{100}

约分:

2

= \frac{- 3 + 7 41}{50}

= - \frac{7 41 - 3}{50}

= - \frac{- 3 + 7 41}{50}

u = \frac{- 6 - 14 41}{2 ( - 50 )} : \frac{3 + 7 41}{50}

\frac{- 6 - 14 41}{2 ( - 50 )}

去除括号:

(- a) = - a

= \frac{- 6 - 14 41}{- 2 \cdot 50}

数字相乘:

2 \cdot 50 = 100

= \frac{- 6 - 14 41}{- 100}

使用分式法则:

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

- 6 - 1441 = - (6 + 1441)

= \frac{6 + 14 41}{100}

分解

6 + 1441 : 2 (3 + 741)

6 + 1441

改写为

= 2 \cdot 3 + 2 \cdot 741

因式分解出通项

2

= 2 (3 + 741)

= \frac{2 ( 3 + 7 41 )}{100}

约分:

2

= \frac{3 + 7 41}{50}

二次方程组的解是:

u = - \frac{- 3 + 7 41}{50}, u = \frac{3 + 7 41}{50}

u = cos (x)

代回

cos (x) = - \frac{- 3 + 7 41}{50}, cos (x) = \frac{3 + 7 41}{50}

cos (x) = - \frac{- 3 + 7 41}{50}, cos (x) = \frac{3 + 7 41}{50}

cos (x) = - \frac{- 3 + 7 41}{50} : x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

cos (x) = - \frac{- 3 + 7 41}{50}

使用反三角函数性质

cos (x) = - \frac{- 3 + 7 41}{50}

cos (x) = - \frac{- 3 + 7 41}{50}

的通解

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

x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

cos (x) = \frac{3 + 7 41}{50} : x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

cos (x) = \frac{3 + 7 41}{50}

使用反三角函数性质

cos (x) = \frac{3 + 7 41}{50}

cos (x) = \frac{3 + 7 41}{50}

的通解

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

x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

合并所有解

x = arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = arccos (\frac{3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

将解代入原方程进行验证

将它们代入

7 sin (x) = cos (x) - 3

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

的解:

假

arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

代入

n = 1

arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 3

代入

x = arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1

7 sin (arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1) = cos (arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1) - 3

整理后得

3.83643 \dots = - 3.83643 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

的解:

真

- arccos (- \frac{- 3 + 7 41}{50}) + 2 πn

代入

n = 1

- arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 3

代入

x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1

7 sin (- arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1) = cos (- arccos (- \frac{- 3 + 7 41}{50}) + 2 π 1) - 3

整理后得

- 3.83643 \dots = - 3.83643 \dots

\Rightarrow 真

检验

arccos (\frac{3 + 7 41}{50}) + 2 πn

的解:

假

arccos (\frac{3 + 7 41}{50}) + 2 πn

代入

n = 1

arccos (\frac{3 + 7 41}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 3

代入

x = arccos (\frac{3 + 7 41}{50}) + 2 π 1

7 sin (arccos (\frac{3 + 7 41}{50}) + 2 π 1) = cos (arccos (\frac{3 + 7 41}{50}) + 2 π 1) - 3

整理后得

2.04356 \dots = - 2.04356 \dots

\Rightarrow 假

检验

2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

的解:

真

2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

代入

n = 1

2 π - arccos (\frac{3 + 7 41}{50}) + 2 π 1

对于

7 sin (x) = cos (x) - 3

代入

x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 π 1

7 sin (2 π - arccos (\frac{3 + 7 41}{50}) + 2 π 1) = cos (2 π - arccos (\frac{3 + 7 41}{50}) + 2 π 1) - 3

整理后得

- 2.04356 \dots = - 2.04356 \dots

\Rightarrow 真

x = - arccos (- \frac{- 3 + 7 41}{50}) + 2 πn, x = 2 π - arccos (\frac{3 + 7 41}{50}) + 2 πn

以小数形式表示解

x = - 2.56154 \dots + 2 πn, x = 2 π - 0.29625 \dots + 2 πn

作图

流行的例子

sin (a) = \frac{3}{4}

sin(7x)+sin(x)=0,0<= x<= 2pi

sin (7 x) + sin (x) = 0, 0 \leq x \leq 2 π

solvefor t,-2sin(t)+2cos(2t)=0

so l v e f or t, - 2 sin (t) + 2 cos (2 t) = 0

2cos(x)-sqrt(3)sin(x)=1

2 cos (x) - 3 sin (x) = 1

cos^2(x)=tan(x)

cos^{2} (x) = tan (x)