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

解答

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

解答

x = - 0.92729 \dots + 2 πn, x = π + 2 πn

+1

度数

x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2 sin (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

用替代法求解

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

令

: sin (x) = u

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

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

- 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

使用指数法则:

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

若

n

是偶数

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

= 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:

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 )} : - \frac{4}{5}

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

去除括号:

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

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

数字相加:

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}

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

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

去除括号:

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

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

数字相减:

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

二次方程组的解是:

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

u = sin (x)

代回

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

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

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

sin (x) = - \frac{4}{5}

使用反三角函数性质

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{4}{5}) + 2 πn

的解:

真

arcsin (- \frac{4}{5}) + 2 πn

代入

n = 1

arcsin (- \frac{4}{5}) + 2 π 1

对于

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

代入

x = arcsin (- \frac{4}{5}) + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

检验

π + arcsin (\frac{4}{5}) + 2 πn

的解:

假

π + arcsin (\frac{4}{5}) + 2 πn

代入

n = 1

π + arcsin (\frac{4}{5}) + 2 π 1

对于

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

代入

x = π + arcsin (\frac{4}{5}) + 2 π 1

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

整理后得

- 2.2 = - 1

\Rightarrow 假

检验

2 πn

的解:

假

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

cos (2 π 1) + 2 sin (2 π 1) = - 1

整理后得

1 = - 1

\Rightarrow 假

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

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

以小数形式表示解

x = - 0.92729 \dots + 2 πn, x = π + 2 πn

流行的例子

csc (x) = - 7

cos(4t)+sin(4t)=0

cos (4 t) + sin (4 t) = 0

cos (x - 3 0^{\circ}) = 0.22

sin(4x-22)=cos(6x-13)

sin (4 x - 22) = cos (6 x - 13)

cos (2 x) + 1 = 2