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受欢迎的三角函数 >

3sin(x)-cos(x)=1

解答

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

解答

x = π + 2 πn, x = 0.64350 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 36.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

- 1 - cos^{2} (x) - 2 cos (x) + 9 sin^{2} (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)

= - 1 - cos^{2} (x) - 2 cos (x) + 9 (1 - cos^{2} (x))

化简

- 1 - cos^{2} (x) - 2 cos (x) + 9 (1 - cos^{2} (x)) : - 10 cos^{2} (x) - 2 cos (x) + 8

- 1 - cos^{2} (x) - 2 cos (x) + 9 (1 - cos^{2} (x))

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

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

= - 1 - cos^{2} (x) - 2 cos (x) + 9 - 9 cos^{2} (x)

化简

- 1 - cos^{2} (x) - 2 cos (x) + 9 - 9 cos^{2} (x) : - 10 cos^{2} (x) - 2 cos (x) + 8

- 1 - cos^{2} (x) - 2 cos (x) + 9 - 9 cos^{2} (x)

对同类项分组

= - cos^{2} (x) - 2 cos (x) - 9 cos^{2} (x) - 1 + 9

同类项相加:

- cos^{2} (x) - 9 cos^{2} (x) = - 10 cos^{2} (x)

= - 10 cos^{2} (x) - 2 cos (x) - 1 + 9

数字相加/相减:

- 1 + 9 = 8

= - 10 cos^{2} (x) - 2 cos (x) + 8

= - 10 cos^{2} (x) - 2 cos (x) + 8

= - 10 cos^{2} (x) - 2 cos (x) + 8

8 - 10 cos^{2} (x) - 2 cos (x) = 0

用替代法求解

8 - 10 cos^{2} (x) - 2 cos (x) = 0

令

: cos (x) = u

8 - 10 u^{2} - 2 u = 0

8 - 10 u^{2} - 2 u = 0 : u = - 1, u = \frac{4}{5}

8 - 10 u^{2} - 2 u = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 10, b = - 2, c = 8

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

(- 2)^{2} - 4 (- 10) \cdot 8 = 18

(- 2)^{2} - 4 (- 10) \cdot 8

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 10 \cdot 8

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 10 \cdot 8

数字相乘:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

数字相加:

4 + 320 = 324

= 324

因式分解数字:

324 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 2 ) \pm 18}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 18}{2 ( - 10 )}, u_{2} = \frac{- ( - 2 ) - 18}{2 ( - 10 )}

u = \frac{- ( - 2 ) + 18}{2 ( - 10 )} : - 1

\frac{- ( - 2 ) + 18}{2 ( - 10 )}

去除括号:

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

= \frac{2 + 18}{- 2 \cdot 10}

数字相加:

2 + 18 = 20

= \frac{20}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{20}{- 20}

使用分式法则:

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

= - \frac{20}{20}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 18}{2 ( - 10 )} : \frac{4}{5}

\frac{- ( - 2 ) - 18}{2 ( - 10 )}

去除括号:

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

= \frac{2 - 18}{- 2 \cdot 10}

数字相减:

2 - 18 = - 16

= \frac{- 16}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 16}{- 20}

使用分式法则:

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

= \frac{16}{20}

约分:

4

= \frac{4}{5}

二次方程组的解是:

u = - 1, u = \frac{4}{5}

u = cos (x)

代回

cos (x) = - 1, cos (x) = \frac{4}{5}

cos (x) = - 1, cos (x) = \frac{4}{5}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

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

cos (x) = \frac{4}{5}

使用反三角函数性质

cos (x) = \frac{4}{5}

cos (x) = \frac{4}{5}

的通解

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

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

3 sin (π + 2 π 1) - cos (π + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

检验

arccos (\frac{4}{5}) + 2 πn

的解:

真

arccos (\frac{4}{5}) + 2 πn

代入

n = 1

arccos (\frac{4}{5}) + 2 π 1

对于

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

代入

x = arccos (\frac{4}{5}) + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

2 π - arccos (\frac{4}{5}) + 2 πn

的解:

假

2 π - arccos (\frac{4}{5}) + 2 πn

代入

n = 1

2 π - arccos (\frac{4}{5}) + 2 π 1

对于

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

代入

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

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

整理后得

- 2.6 = 1

\Rightarrow 假

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

以小数形式表示解

x = π + 2 πn, x = 0.64350 \dots + 2 πn

作图

流行的例子

tan (x) = - 0.73

cos(x)=2sin^2(x)

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

cos^2(x)-sin^2(x)=2-5cos(x)

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

tan(x/4)=-sqrt(3)

tan (\frac{x}{4}) = - 3

8sin^2(x)-2sin(x)-1=0,-pi/2 <x< pi/2

8 sin^{2} (x) - 2 sin (x) - 1 = 0, - \frac{π}{2} < x < \frac{π}{2}