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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = 2 π - 0.64350 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 323.13010 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2 cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

= 4 cos (x) - 5 cos^{2} (x)

4 cos (x) - 5 cos^{2} (x) = 0

用替代法求解

4 cos (x) - 5 cos^{2} (x) = 0

令

: cos (x) = u

4 u - 5 u^{2} = 0

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

4 u - 5 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 5, b = 4, c = 0

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

4^{2} - 4 (- 5) \cdot 0 = 4

4^{2} - 4 (- 5) \cdot 0

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 5 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} + 0

4^{2} + 0 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

u_{1, 2} = \frac{- 4 \pm 4}{2 ( - 5 )}

将解分隔开

u_{1} = \frac{- 4 + 4}{2 ( - 5 )}, u_{2} = \frac{- 4 - 4}{2 ( - 5 )}

u = \frac{- 4 + 4}{2 ( - 5 )} : 0

\frac{- 4 + 4}{2 ( - 5 )}

去除括号:

(- a) = - a

= \frac{- 4 + 4}{- 2 \cdot 5}

数字相加/相减:

- 4 + 4 = 0

= \frac{0}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{0}{- 10}

使用分式法则:

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

= - \frac{0}{10}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

\frac{- 4 - 4}{2 ( - 5 )}

去除括号:

(- a) = - a

= \frac{- 4 - 4}{- 2 \cdot 5}

数字相减:

- 4 - 4 = - 8

= \frac{- 8}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 8}{- 10}

使用分式法则:

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

= \frac{8}{10}

约分:

2

= \frac{4}{5}

二次方程组的解是:

u = 0, u = \frac{4}{5}

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{4}{5}

cos (x) = 0, cos (x) = \frac{4}{5}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 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

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

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

sin (\frac{3 π}{2} + 2 π 1) + 2 cos (\frac{3 π}{2} + 2 π 1) = 1

整理后得

- 1 = 1

\Rightarrow 假

检验

arccos (\frac{4}{5}) + 2 πn

的解:

假

arccos (\frac{4}{5}) + 2 πn

代入

n = 1

arccos (\frac{4}{5}) + 2 π 1

对于

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

代入

x = arccos (\frac{4}{5}) + 2 π 1

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

整理后得

2.2 = 1

\Rightarrow 假

检验

2 π - arccos (\frac{4}{5}) + 2 πn

的解:

真

2 π - arccos (\frac{4}{5}) + 2 πn

代入

n = 1

2 π - arccos (\frac{4}{5}) + 2 π 1

对于

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

代入

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

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

整理后得

1 = 1

\Rightarrow 真

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

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = 2 π - 0.64350 \dots + 2 πn

作图

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