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受欢迎的三角函数 >

5sin(x)=cos(x)-4

解答

5 sin (x) = cos (x) - 4

解答

x = - 2.04236 \dots + 2 πn, x = 2 π - 0.70443 \dots + 2 πn

+1

度数

x = - 117.01888 \dots^{\circ} + 36 0^{\circ} n, x = 319.63875 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 sin (x) = cos (x) - 4

两边进行平方

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

两边减去

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

25 sin^{2} (x) - cos^{2} (x) + 8 cos (x) - 16 = 0

使用三角恒等式改写

- 16 - cos^{2} (x) + 25 sin^{2} (x) + 8 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)

= - 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x)

化简

- 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 26 cos^{2} (x) + 9

- 16 - cos^{2} (x) + 25 (1 - cos^{2} (x)) + 8 cos (x)

乘开

25 (1 - cos^{2} (x)) : 25 - 25 cos^{2} (x)

25 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 25, b = 1, c = cos^{2} (x)

= 25 \cdot 1 - 25 cos^{2} (x)

数字相乘:

25 \cdot 1 = 25

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

= - 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x)

化简

- 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x) : 8 cos (x) - 26 cos^{2} (x) + 9

- 16 - cos^{2} (x) + 25 - 25 cos^{2} (x) + 8 cos (x)

对同类项分组

= - cos^{2} (x) - 25 cos^{2} (x) + 8 cos (x) - 16 + 25

同类项相加:

- cos^{2} (x) - 25 cos^{2} (x) = - 26 cos^{2} (x)

= - 26 cos^{2} (x) + 8 cos (x) - 16 + 25

数字相加/相减:

- 16 + 25 = 9

= 8 cos (x) - 26 cos^{2} (x) + 9

= 8 cos (x) - 26 cos^{2} (x) + 9

= 8 cos (x) - 26 cos^{2} (x) + 9

9 - 26 cos^{2} (x) + 8 cos (x) = 0

用替代法求解

9 - 26 cos^{2} (x) + 8 cos (x) = 0

令

: cos (x) = u

9 - 26 u^{2} + 8 u = 0

9 - 26 u^{2} + 8 u = 0 : u = - \frac{- 4 + 5 10}{26}, u = \frac{4 + 5 10}{26}

9 - 26 u^{2} + 8 u = 0

改写成标准形式

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

- 26 u^{2} + 8 u + 9 = 0

使用求根公式求解

- 26 u^{2} + 8 u + 9 = 0

二次方程求根公式:

若

a = - 26, b = 8, c = 9

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 26 ) \cdot 9}{2 ( - 26 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 26 ) \cdot 9}{2 ( - 26 )}

8^{2} - 4 (- 26) \cdot 9 = 1010

8^{2} - 4 (- 26) \cdot 9

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 26 \cdot 9

数字相乘:

4 \cdot 26 \cdot 9 = 936

= 8^{2} + 936

8^{2} = 64

= 64 + 936

数字相加:

64 + 936 = 1000

= 1000

1000

质因数分解:

2^{3} \cdot 5^{3}

1000

1000

除以

21000 = 500 \cdot 2

= 2 \cdot 500

500

除以

2500 = 250 \cdot 2

= 2 \cdot 2 \cdot 250

250

除以

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 125

125

除以

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

= 2^{3} \cdot 5^{3}

= 2^{3} \cdot 5^{3}

使用指数法则:

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

= 2^{2} \cdot 5^{2} \cdot 2 \cdot 5

使用根式运算法则:

n ab = n a n b

= 2^{2} 5^{2} 2 \cdot 5

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 5^{2} 2 \cdot 5

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 2 \cdot 5 2 \cdot 5

整理后得

= 1010

u_{1, 2} = \frac{- 8 \pm 10 10}{2 ( - 26 )}

将解分隔开

u_{1} = \frac{- 8 + 10 10}{2 ( - 26 )}, u_{2} = \frac{- 8 - 10 10}{2 ( - 26 )}

u = \frac{- 8 + 10 10}{2 ( - 26 )} : - \frac{- 4 + 5 10}{26}

\frac{- 8 + 10 10}{2 ( - 26 )}

去除括号:

(- a) = - a

= \frac{- 8 + 10 10}{- 2 \cdot 26}

数字相乘:

2 \cdot 26 = 52

= \frac{- 8 + 10 10}{- 52}

使用分式法则:

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

= - \frac{- 8 + 10 10}{52}

消掉

\frac{- 8 + 10 10}{52} : \frac{5 10 - 4}{26}

\frac{- 8 + 10 10}{52}

分解

- 8 + 1010 : 2 (- 4 + 510)

- 8 + 1010

改写为

= - 2 \cdot 4 + 2 \cdot 510

因式分解出通项

2

= 2 (- 4 + 510)

= \frac{2 ( - 4 + 5 10 )}{52}

约分:

2

= \frac{- 4 + 5 10}{26}

= - \frac{5 10 - 4}{26}

= - \frac{- 4 + 5 10}{26}

u = \frac{- 8 - 10 10}{2 ( - 26 )} : \frac{4 + 5 10}{26}

\frac{- 8 - 10 10}{2 ( - 26 )}

去除括号:

(- a) = - a

= \frac{- 8 - 10 10}{- 2 \cdot 26}

数字相乘:

2 \cdot 26 = 52

= \frac{- 8 - 10 10}{- 52}

使用分式法则:

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

- 8 - 1010 = - (8 + 1010)

= \frac{8 + 10 10}{52}

分解

8 + 1010 : 2 (4 + 510)

8 + 1010

改写为

= 2 \cdot 4 + 2 \cdot 510

因式分解出通项

2

= 2 (4 + 510)

= \frac{2 ( 4 + 5 10 )}{52}

约分:

2

= \frac{4 + 5 10}{26}

二次方程组的解是:

u = - \frac{- 4 + 5 10}{26}, u = \frac{4 + 5 10}{26}

u = cos (x)

代回

cos (x) = - \frac{- 4 + 5 10}{26}, cos (x) = \frac{4 + 5 10}{26}

cos (x) = - \frac{- 4 + 5 10}{26}, cos (x) = \frac{4 + 5 10}{26}

cos (x) = - \frac{- 4 + 5 10}{26} : x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

cos (x) = - \frac{- 4 + 5 10}{26}

使用反三角函数性质

cos (x) = - \frac{- 4 + 5 10}{26}

cos (x) = - \frac{- 4 + 5 10}{26}

的通解

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

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

cos (x) = \frac{4 + 5 10}{26} : x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

cos (x) = \frac{4 + 5 10}{26}

使用反三角函数性质

cos (x) = \frac{4 + 5 10}{26}

cos (x) = \frac{4 + 5 10}{26}

的通解

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

x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

合并所有解

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = arccos (\frac{4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

将解代入原方程进行验证

将它们代入

5 sin (x) = cos (x) - 4

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

的解:

假

arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

代入

n = 1

arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

对于

5 sin (x) = cos (x) - 4

代入

x = arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

5 sin (arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) = cos (arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) - 4

整理后得

4.45428 \dots = - 4.45428 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

的解:

真

- arccos (- \frac{- 4 + 5 10}{26}) + 2 πn

代入

n = 1

- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

对于

5 sin (x) = cos (x) - 4

代入

x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1

5 sin (- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) = cos (- arccos (- \frac{- 4 + 5 10}{26}) + 2 π 1) - 4

整理后得

- 4.45428 \dots = - 4.45428 \dots

\Rightarrow 真

检验

arccos (\frac{4 + 5 10}{26}) + 2 πn

的解:

假

arccos (\frac{4 + 5 10}{26}) + 2 πn

代入

n = 1

arccos (\frac{4 + 5 10}{26}) + 2 π 1

对于

5 sin (x) = cos (x) - 4

代入

x = arccos (\frac{4 + 5 10}{26}) + 2 π 1

5 sin (arccos (\frac{4 + 5 10}{26}) + 2 π 1) = cos (arccos (\frac{4 + 5 10}{26}) + 2 π 1) - 4

整理后得

3.23802 \dots = - 3.23802 \dots

\Rightarrow 假

检验

2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

的解:

真

2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

代入

n = 1

2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1

对于

5 sin (x) = cos (x) - 4

代入

x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1

5 sin (2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1) = cos (2 π - arccos (\frac{4 + 5 10}{26}) + 2 π 1) - 4

整理后得

- 3.23802 \dots = - 3.23802 \dots

\Rightarrow 真

x = - arccos (- \frac{- 4 + 5 10}{26}) + 2 πn, x = 2 π - arccos (\frac{4 + 5 10}{26}) + 2 πn

以小数形式表示解

x = - 2.04236 \dots + 2 πn, x = 2 π - 0.70443 \dots + 2 πn

作图

流行的例子

-21tan(x)+3sqrt(3)=-3(sqrt(3)+tan(x))

- 21 tan (x) + 33 = - 3 (3 + tan (x))

tan (y) = - 1

tan (x) = \frac{12}{4}

1-sqrt(2)cos(θ)=0

1 - 2 cos (θ) = 0

sqrt(3)tan(3x)-1=0,0<= x<= 2pi

3 tan (3 x) - 1 = 0, 0 \leq x \leq 2 π