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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 306.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cos (x)

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

两边进行平方

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

两边减去

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

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

化简

- 1 - cos^{2} (x) - 2 cos (x) + 4 (1 - cos^{2} (x)) : - 5 cos^{2} (x) - 2 cos (x) + 3

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

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

- 1 - cos^{2} (x) - 2 cos (x) + 4 - 4 cos^{2} (x) : - 5 cos^{2} (x) - 2 cos (x) + 3

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

对同类项分组

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

同类项相加:

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

= - 5 cos^{2} (x) - 2 cos (x) - 1 + 4

数字相加/相减:

- 1 + 4 = 3

= - 5 cos^{2} (x) - 2 cos (x) + 3

= - 5 cos^{2} (x) - 2 cos (x) + 3

= - 5 cos^{2} (x) - 2 cos (x) + 3

3 - 2 cos (x) - 5 cos^{2} (x) = 0

用替代法求解

3 - 2 cos (x) - 5 cos^{2} (x) = 0

令

: cos (x) = u

3 - 2 u - 5 u^{2} = 0

3 - 2 u - 5 u^{2} = 0 : u = - 1, u = \frac{3}{5}

3 - 2 u - 5 u^{2} = 0

改写成标准形式

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

- 5 u^{2} - 2 u + 3 = 0

使用求根公式求解

- 5 u^{2} - 2 u + 3 = 0

二次方程求根公式:

若

a = - 5, b = - 2, c = 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

(- 2)^{2} - 4 (- 5) \cdot 3 = 8

(- 2)^{2} - 4 (- 5) \cdot 3

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 5 \cdot 3

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 5 \cdot 3

数字相乘:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

数字相加:

4 + 60 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 ( - 5 )}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 8}{2 ( - 5 )}, u_{2} = \frac{- ( - 2 ) - 8}{2 ( - 5 )}

u = \frac{- ( - 2 ) + 8}{2 ( - 5 )} : - 1

\frac{- ( - 2 ) + 8}{2 ( - 5 )}

去除括号:

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

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

数字相加:

2 + 8 = 10

= \frac{10}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10}{- 10}

使用分式法则:

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

= - \frac{10}{10}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 8}{2 ( - 5 )} : \frac{3}{5}

\frac{- ( - 2 ) - 8}{2 ( - 5 )}

去除括号:

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

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

数字相减:

2 - 8 = - 6

= \frac{- 6}{- 2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 6}{- 10}

使用分式法则:

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

= \frac{6}{10}

约分:

2

= \frac{3}{5}

二次方程组的解是:

u = - 1, u = \frac{3}{5}

u = cos (x)

代回

cos (x) = - 1, cos (x) = \frac{3}{5}

cos (x) = - 1, cos (x) = \frac{3}{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{3}{5} : x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (x) = \frac{3}{5}

使用反三角函数性质

cos (x) = \frac{3}{5}

cos (x) = \frac{3}{5}

的通解

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

x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

合并所有解

x = π + 2 πn, x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

将解代入原方程进行验证

将它们代入

2 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

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

代入

x = π + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

检验

arccos (\frac{3}{5}) + 2 πn

的解:

假

arccos (\frac{3}{5}) + 2 πn

代入

n = 1

arccos (\frac{3}{5}) + 2 π 1

对于

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

代入

x = arccos (\frac{3}{5}) + 2 π 1

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

整理后得

2.2 = - 1

\Rightarrow 假

检验

2 π - arccos (\frac{3}{5}) + 2 πn

的解:

真

2 π - arccos (\frac{3}{5}) + 2 πn

代入

n = 1

2 π - arccos (\frac{3}{5}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{3}{5}) + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

x = π + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

以小数形式表示解

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

作图

流行的例子

sec^2(x)+7tan(x)=9

sec^{2} (x) + 7 tan (x) = 9

1-3cos(θ)=sin^2(θ)

1 - 3 cos (θ) = sin^{2} (θ)

cot (2 x) = - 1

2 sec (θ) + 3 = 0

3sec^2(x)-5tan(x)=3

3 sec^{2} (x) - 5 tan (x) = 3