zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

2sin(x)+2cos(x)=sqrt(6)

解答

2 sin (x) + 2 cos (x) = 6

解答

x = 1.30899 \dots + 2 πn, x = 0.26179 \dots + 2 πn

+1

度数

x = 7 5^{\circ} + 36 0^{\circ} n, x = 1 5^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x) + 2 cos (x) = 6

两边减去

2 cos (x)

2 sin (x) = 6 - 2 cos (x)

两边进行平方

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

两边减去

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

4 sin^{2} (x) - 6 + 46 cos (x) - 4 cos^{2} (x) = 0

使用三角恒等式改写

- 6 - 4 cos^{2} (x) + 4 sin^{2} (x) + 4 cos (x) 6

使用毕达哥拉斯恒等式:

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

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

= - 6 - 4 cos^{2} (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) 6

化简

- 6 - 4 cos^{2} (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) 6 : 46 cos (x) - 8 cos^{2} (x) - 2

- 6 - 4 cos^{2} (x) + 4 (1 - cos^{2} (x)) + 4 cos (x) 6

= - 6 - 4 cos^{2} (x) + 4 (1 - cos^{2} (x)) + 46 cos (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)

= - 6 - 4 cos^{2} (x) + 4 - 4 cos^{2} (x) + 4 cos (x) 6

化简

- 6 - 4 cos^{2} (x) + 4 - 4 cos^{2} (x) + 4 cos (x) 6 : 46 cos (x) - 8 cos^{2} (x) - 2

- 6 - 4 cos^{2} (x) + 4 - 4 cos^{2} (x) + 4 cos (x) 6

对同类项分组

= - 4 cos^{2} (x) - 4 cos^{2} (x) + 46 cos (x) - 6 + 4

同类项相加:

- 4 cos^{2} (x) - 4 cos^{2} (x) = - 8 cos^{2} (x)

= - 8 cos^{2} (x) + 46 cos (x) - 6 + 4

数字相加/相减:

- 6 + 4 = - 2

= 46 cos (x) - 8 cos^{2} (x) - 2

= 46 cos (x) - 8 cos^{2} (x) - 2

= 46 cos (x) - 8 cos^{2} (x) - 2

- 2 - 8 cos^{2} (x) + 4 cos (x) 6 = 0

用替代法求解

- 2 - 8 cos^{2} (x) + 4 cos (x) 6 = 0

令

: cos (x) = u

- 2 - 8 u^{2} + 4 u 6 = 0

- 2 - 8 u^{2} + 4 u 6 = 0 : u = \frac{6 - 2}{4}, u = \frac{6 + 2}{4}

- 2 - 8 u^{2} + 4 u 6 = 0

改写成标准形式

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

- 8 u^{2} + 46 u - 2 = 0

使用求根公式求解

- 8 u^{2} + 46 u - 2 = 0

二次方程求根公式:

若

a = - 8, b = 46, c = - 2

u_{1, 2} = \frac{- 4 6 \pm ( 4 6 ) ^{2} - 4 ( - 8 ) ( - 2 )}{2 ( - 8 )}

u_{1, 2} = \frac{- 4 6 \pm ( 4 6 ) ^{2} - 4 ( - 8 ) ( - 2 )}{2 ( - 8 )}

(46)^{2} - 4 (- 8) (- 2) = 42

(46)^{2} - 4 (- 8) (- 2)

使用法则

- (- a) = a

= (46)^{2} - 4 \cdot 8 \cdot 2

(46)^{2} = 4^{2} \cdot 6

(46)^{2}

使用指数法则:

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

= 4^{2} (6)^{2}

(6)^{2} : 6

使用根式运算法则:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 6

= 4^{2} \cdot 6

4 \cdot 8 \cdot 2 = 64

4 \cdot 8 \cdot 2

数字相乘:

4 \cdot 8 \cdot 2 = 64

= 64

= 4^{2} \cdot 6 - 64

4^{2} \cdot 6 = 96

4^{2} \cdot 6

4^{2} = 16

= 16 \cdot 6

数字相乘:

16 \cdot 6 = 96

= 96

= 96 - 64

数字相减:

96 - 64 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

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

2

是质数，因此无法进一步因数分解

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

= 2^{5}

= 2^{5}

使用指数法则:

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

= 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 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} 2

整理后得

= 42

u_{1, 2} = \frac{- 4 6 \pm 4 2}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 4 6 + 4 2}{2 ( - 8 )}, u_{2} = \frac{- 4 6 - 4 2}{2 ( - 8 )}

u = \frac{- 4 6 + 4 2}{2 ( - 8 )} : \frac{6 - 2}{4}

\frac{- 4 6 + 4 2}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 4 6 + 4 2}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4 6 + 4 2}{- 16}

使用分式法则:

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

- 46 + 42 = - (46 - 42)

= \frac{4 6 - 4 2}{16}

因式分解出通项

4

= \frac{4 ( 6 - 2 )}{16}

约分:

4

= \frac{6 - 2}{4}

u = \frac{- 4 6 - 4 2}{2 ( - 8 )} : \frac{6 + 2}{4}

\frac{- 4 6 - 4 2}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 4 6 - 4 2}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4 6 - 4 2}{- 16}

使用分式法则:

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

- 46 - 42 = - (46 + 42)

= \frac{4 6 + 4 2}{16}

因式分解出通项

4

= \frac{4 ( 6 + 2 )}{16}

约分:

4

= \frac{6 + 2}{4}

二次方程组的解是:

u = \frac{6 - 2}{4}, u = \frac{6 + 2}{4}

u = cos (x)

代回

cos (x) = \frac{6 - 2}{4}, cos (x) = \frac{6 + 2}{4}

cos (x) = \frac{6 - 2}{4}, cos (x) = \frac{6 + 2}{4}

cos (x) = \frac{6 - 2}{4} : x = arccos (\frac{6 - 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 - 2}{4}) + 2 πn

cos (x) = \frac{6 - 2}{4}

使用反三角函数性质

cos (x) = \frac{6 - 2}{4}

cos (x) = \frac{6 - 2}{4}

的通解

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

x = arccos (\frac{6 - 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 - 2}{4}) + 2 πn

x = arccos (\frac{6 - 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 - 2}{4}) + 2 πn

cos (x) = \frac{6 + 2}{4} : x = arccos (\frac{6 + 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 + 2}{4}) + 2 πn

cos (x) = \frac{6 + 2}{4}

使用反三角函数性质

cos (x) = \frac{6 + 2}{4}

cos (x) = \frac{6 + 2}{4}

的通解

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

x = arccos (\frac{6 + 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 + 2}{4}) + 2 πn

x = arccos (\frac{6 + 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 + 2}{4}) + 2 πn

合并所有解

x = arccos (\frac{6 - 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 - 2}{4}) + 2 πn, x = arccos (\frac{6 + 2}{4}) + 2 πn, x = 2 π - arccos (\frac{6 + 2}{4}) + 2 πn

将解代入原方程进行验证

将它们代入

2 sin (x) + 2 cos (x) = 6

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{6 - 2}{4}) + 2 πn

的解:

真

arccos (\frac{6 - 2}{4}) + 2 πn

代入

n = 1

arccos (\frac{6 - 2}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 6

代入

x = arccos (\frac{6 - 2}{4}) + 2 π 1

2 sin (arccos (\frac{6 - 2}{4}) + 2 π 1) + 2 cos (arccos (\frac{6 - 2}{4}) + 2 π 1) = 6

整理后得

2.44948 \dots = 2.44948 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{6 - 2}{4}) + 2 πn

的解:

假

2 π - arccos (\frac{6 - 2}{4}) + 2 πn

代入

n = 1

2 π - arccos (\frac{6 - 2}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 6

代入

x = 2 π - arccos (\frac{6 - 2}{4}) + 2 π 1

2 sin (2 π - arccos (\frac{6 - 2}{4}) + 2 π 1) + 2 cos (2 π - arccos (\frac{6 - 2}{4}) + 2 π 1) = 6

整理后得

- 1.41421 \dots = 2.44948 \dots

\Rightarrow 假

检验

arccos (\frac{6 + 2}{4}) + 2 πn

的解:

真

arccos (\frac{6 + 2}{4}) + 2 πn

代入

n = 1

arccos (\frac{6 + 2}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 6

代入

x = arccos (\frac{6 + 2}{4}) + 2 π 1

2 sin (arccos (\frac{6 + 2}{4}) + 2 π 1) + 2 cos (arccos (\frac{6 + 2}{4}) + 2 π 1) = 6

整理后得

2.44948 \dots = 2.44948 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{6 + 2}{4}) + 2 πn

的解:

假

2 π - arccos (\frac{6 + 2}{4}) + 2 πn

代入

n = 1

2 π - arccos (\frac{6 + 2}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 6

代入

x = 2 π - arccos (\frac{6 + 2}{4}) + 2 π 1

2 sin (2 π - arccos (\frac{6 + 2}{4}) + 2 π 1) + 2 cos (2 π - arccos (\frac{6 + 2}{4}) + 2 π 1) = 6

整理后得

1.41421 \dots = 2.44948 \dots

\Rightarrow 假

x = arccos (\frac{6 - 2}{4}) + 2 πn, x = arccos (\frac{6 + 2}{4}) + 2 πn

以小数形式表示解

x = 1.30899 \dots + 2 πn, x = 0.26179 \dots + 2 πn

作图

流行的例子

0=-0.65sin(13x)

0 = - 0.65 sin (13 x)

solvefor x,tan^2(x)-1=0

so l v e f or x, tan^{2} (x) - 1 = 0

c^9=3tan(-3x+c)

c^{9} = 3 tan (- 3 x + c)

1 = 8 + 7 sin (15 t)

4sin(θ)=4-4sin(θ)

4 sin (θ) = 4 - 4 sin (θ)