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受欢迎的三角函数 >

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

解答

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

解答

x = 2.87979 \dots + 2 πn, x = 1.83259 \dots + 2 πn

+1

度数

x = 16 5^{\circ} + 36 0^{\circ} n, x = 10 5^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

2 cos (x)

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

两边进行平方

(2 sin (x))^{2} = (3 + 2 cos (x))^{2}

两边减去

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

2 sin^{2} (x) - 3 - 26 cos (x) - 2 cos^{2} (x) = 0

使用三角恒等式改写

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

= - 3 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 2 cos (x) 6

化简

- 3 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 2 cos (x) 6 : - 4 cos^{2} (x) - 26 cos (x) - 1

- 3 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 2 cos (x) 6

= - 3 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 26 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)

= - 3 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 2 cos (x) 6

化简

- 3 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 2 cos (x) 6 : - 4 cos^{2} (x) - 26 cos (x) - 1

- 3 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 2 cos (x) 6

对同类项分组

= - 2 cos^{2} (x) - 2 cos^{2} (x) - 26 cos (x) - 3 + 2

同类项相加:

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

= - 4 cos^{2} (x) - 26 cos (x) - 3 + 2

数字相加/相减:

- 3 + 2 = - 1

= - 4 cos^{2} (x) - 26 cos (x) - 1

= - 4 cos^{2} (x) - 26 cos (x) - 1

= - 4 cos^{2} (x) - 26 cos (x) - 1

- 1 - 4 cos^{2} (x) - 2 cos (x) 6 = 0

用替代法求解

- 1 - 4 cos^{2} (x) - 2 cos (x) 6 = 0

令

: cos (x) = u

- 1 - 4 u^{2} - 2 u 6 = 0

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

- 1 - 4 u^{2} - 2 u 6 = 0

改写成标准形式

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

- 4 u^{2} - 26 u - 1 = 0

使用求根公式求解

- 4 u^{2} - 26 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = - 26, c = - 1

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

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

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

(- 26)^{2} - 4 (- 4) (- 1)

使用法则

- (- a) = a

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

(- 26)^{2} = 2^{2} \cdot 6

(- 26)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 26)^{2} = (26)^{2}

= (26)^{2}

使用指数法则:

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

= 2^{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

= 2^{2} \cdot 6

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{2} \cdot 6 - 16

2^{2} \cdot 6 = 24

2^{2} \cdot 6

2^{2} = 4

= 4 \cdot 6

数字相乘:

4 \cdot 6 = 24

= 24

= 24 - 16

数字相减:

24 - 16 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 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

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 4 = 8

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

使用分式法则:

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

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

消掉

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

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

因式分解出通项

2

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

约分:

2

= \frac{6 + 2}{4}

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 4 = 8

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

使用分式法则:

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

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

消掉

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

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

因式分解出通项

2

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

约分:

2

= \frac{6 - 2}{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 = - 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 = - arccos (- a) + 2 πn

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

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

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

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

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

合并所有解

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn, x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - 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) = 3

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) = 3

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) = 3

整理后得

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

检验

- arccos (- \frac{6 + 2}{4}) + 2 πn

的解:

假

- 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) = 3

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) = 3

整理后得

1 = 1.73205 \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) = 3

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) = 3

整理后得

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

检验

- arccos (- \frac{6 - 2}{4}) + 2 πn

的解:

假

- 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) = 3

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) = 3

整理后得

- 1 = 1.73205 \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 = 2.87979 \dots + 2 πn, x = 1.83259 \dots + 2 πn

作图

流行的例子

8cos(2x)=4sqrt(3)

8 cos (2 x) = 43

cos^2(θ)-sin^2(θ)=1

cos^{2} (θ) - sin^{2} (θ) = 1

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

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

cos^3(x)=cos^3(x)csc(x)

cos^{3} (x) = cos^{3} (x) csc (x)

- \frac{1}{2} = sin (θ)