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受欢迎的三角函数 >

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

解答

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

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

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

两边减去

3 cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

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

化简

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

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

同类项相加:

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

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

数字相加/相减:

- 2 + 1 = - 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

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

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

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

代入

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

sin (arccos (- \frac{6 + 2}{4}) + 2 π 1) + 3 cos (arccos (- \frac{6 + 2}{4}) + 2 π 1) = - 2

整理后得

- 1.41421 \dots = - 1.41421 \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

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

代入

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

sin (- arccos (- \frac{6 + 2}{4}) + 2 π 1) + 3 cos (- arccos (- \frac{6 + 2}{4}) + 2 π 1) = - 2

整理后得

- 1.93185 \dots = - 1.41421 \dots

\Rightarrow 假

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

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

代入

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

sin (arccos (- \frac{6 - 2}{4}) + 2 π 1) + 3 cos (arccos (- \frac{6 - 2}{4}) + 2 π 1) = - 2

整理后得

0.51763 \dots = - 1.41421 \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

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

代入

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

sin (- arccos (- \frac{6 - 2}{4}) + 2 π 1) + 3 cos (- arccos (- \frac{6 - 2}{4}) + 2 π 1) = - 2

整理后得

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow 真

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

cot(θ)+4csc(θ)=6

cot (θ) + 4 csc (θ) = 6

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

sin (x) 2 = - sin (x)

sin^2(x)-9cos(x)-9=0

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

sin (a) = \frac{8}{17}

cot (t) = 1