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受欢迎的三角函数 >

sqrt(2)sin(x)+sqrt(2)cos(x)=1

解答

2 sin (x) + 2 cos (x) = 1

解答

x = 1.83259 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn

+1

度数

x = 10 5^{\circ} + 36 0^{\circ} n, x = 34 5^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x) + 2 cos (x) = 1

两边减去

2 cos (x)

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

两边进行平方

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

两边减去

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

2 sin^{2} (x) - 1 + 22 cos (x) - 2 cos^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

- 1 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) + 2 cos (x) 2 : 22 cos (x) - 4 cos^{2} (x) + 1

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

= - 1 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) + 22 cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

2 \cdot 1 = 2

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

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

化简

- 1 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) + 2 cos (x) 2 : 22 cos (x) - 4 cos^{2} (x) + 1

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

对同类项分组

= - 2 cos^{2} (x) - 2 cos^{2} (x) + 22 cos (x) - 1 + 2

同类项相加:

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

= - 4 cos^{2} (x) + 22 cos (x) - 1 + 2

数字相加/相减:

- 1 + 2 = 1

= 22 cos (x) - 4 cos^{2} (x) + 1

= 22 cos (x) - 4 cos^{2} (x) + 1

= 22 cos (x) - 4 cos^{2} (x) + 1

1 - 4 cos^{2} (x) + 2 cos (x) 2 = 0

用替代法求解

1 - 4 cos^{2} (x) + 2 cos (x) 2 = 0

令

: cos (x) = u

1 - 4 u^{2} + 2 u 2 = 0

1 - 4 u^{2} + 2 u 2 = 0 : u = - \frac{- 2 + 6}{4}, u = \frac{2 + 6}{4}

1 - 4 u^{2} + 2 u 2 = 0

改写成标准形式

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

- 4 u^{2} + 22 u + 1 = 0

使用求根公式求解

- 4 u^{2} + 22 u + 1 = 0

二次方程求根公式:

若

a = - 4, b = 22, c = 1

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

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

(22)^{2} - 4 (- 4) \cdot 1 = 26

(22)^{2} - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= (22)^{2} + 4 \cdot 4 \cdot 1

(22)^{2} = 2^{3}

(22)^{2}

使用指数法则:

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

= 2^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

= 2^{2} \cdot 2

使用指数法则:

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

2^{2} \cdot 2 = 2^{2 + 1}

= 2^{2 + 1}

数字相加:

2 + 1 = 3

= 2^{3}

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{3} + 16

2^{3} = 8

= 8 + 16

数字相加:

8 + 16 = 24

= 24

24

质因数分解:

2^{3} \cdot 3

24

24

除以

224 = 12 \cdot 2

= 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

使用指数法则:

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

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

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 3

整理后得

= 26

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

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

使用分式法则:

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

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

消掉

\frac{- 2 2 + 2 6}{8} : \frac{6 - 2}{4}

\frac{- 2 2 + 2 6}{8}

因式分解出通项

2

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

约分:

2

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

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

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

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

使用分式法则:

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

- 22 - 26 = - (22 + 26)

= \frac{2 2 + 2 6}{8}

因式分解出通项

2

= \frac{2 ( 2 + 6 )}{8}

约分:

2

= \frac{2 + 6}{4}

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 2 + 6}{4}) + 2 πn

的解:

真

arccos (- \frac{- 2 + 6}{4}) + 2 πn

代入

n = 1

arccos (- \frac{- 2 + 6}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 1

代入

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

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

整理后得

1 = 1

\Rightarrow 真

检验

- arccos (- \frac{- 2 + 6}{4}) + 2 πn

的解:

假

- arccos (- \frac{- 2 + 6}{4}) + 2 πn

代入

n = 1

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

对于

2 sin (x) + 2 cos (x) = 1

代入

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

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

整理后得

- 1.73205 \dots = 1

\Rightarrow 假

检验

arccos (\frac{2 + 6}{4}) + 2 πn

的解:

假

arccos (\frac{2 + 6}{4}) + 2 πn

代入

n = 1

arccos (\frac{2 + 6}{4}) + 2 π 1

对于

2 sin (x) + 2 cos (x) = 1

代入

x = arccos (\frac{2 + 6}{4}) + 2 π 1

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

整理后得

1.73205 \dots = 1

\Rightarrow 假

检验

2 π - arccos (\frac{2 + 6}{4}) + 2 πn

的解:

真

2 π - arccos (\frac{2 + 6}{4}) + 2 πn

代入

n = 1

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

对于

2 sin (x) + 2 cos (x) = 1

代入

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

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

整理后得

1 = 1

\Rightarrow 真

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

以小数形式表示解

x = 1.83259 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn

作图

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