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受欢迎的三角函数 >

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

解答

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

解答

x = 1.17600 \dots + 2 πn, x = 2 πn

+1

度数

x = 67.38013 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3 cos (x)

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

两边进行平方

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

两边减去

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

4 sin^{2} (x) - 9 + 18 cos (x) - 9 cos^{2} (x) = 0

使用三角恒等式改写

- 9 + 18 cos (x) + 4 sin^{2} (x) - 9 cos^{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)

= - 9 + 18 cos (x) + 4 (1 - cos^{2} (x)) - 9 cos^{2} (x)

化简

- 9 + 18 cos (x) + 4 (1 - cos^{2} (x)) - 9 cos^{2} (x) : 18 cos (x) - 13 cos^{2} (x) - 5

- 9 + 18 cos (x) + 4 (1 - cos^{2} (x)) - 9 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)

= - 9 + 18 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x)

化简

- 9 + 18 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x) : 18 cos (x) - 13 cos^{2} (x) - 5

- 9 + 18 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x)

同类项相加:

- 4 cos^{2} (x) - 9 cos^{2} (x) = - 13 cos^{2} (x)

= - 9 + 18 cos (x) + 4 - 13 cos^{2} (x)

对同类项分组

= 18 cos (x) - 13 cos^{2} (x) - 9 + 4

数字相加/相减:

- 9 + 4 = - 5

= 18 cos (x) - 13 cos^{2} (x) - 5

= 18 cos (x) - 13 cos^{2} (x) - 5

= 18 cos (x) - 13 cos^{2} (x) - 5

- 5 - 13 cos^{2} (x) + 18 cos (x) = 0

用替代法求解

- 5 - 13 cos^{2} (x) + 18 cos (x) = 0

令

: cos (x) = u

- 5 - 13 u^{2} + 18 u = 0

- 5 - 13 u^{2} + 18 u = 0 : u = \frac{5}{13}, u = 1

- 5 - 13 u^{2} + 18 u = 0

改写成标准形式

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

- 13 u^{2} + 18 u - 5 = 0

使用求根公式求解

- 13 u^{2} + 18 u - 5 = 0

二次方程求根公式:

若

a = - 13, b = 18, c = - 5

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 13 ) ( - 5 )}{2 ( - 13 )}

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 13 ) ( - 5 )}{2 ( - 13 )}

1 8^{2} - 4 (- 13) (- 5) = 8

1 8^{2} - 4 (- 13) (- 5)

使用法则

- (- a) = a

= 1 8^{2} - 4 \cdot 13 \cdot 5

数字相乘:

4 \cdot 13 \cdot 5 = 260

= 1 8^{2} - 260

1 8^{2} = 324

= 324 - 260

数字相减:

324 - 260 = 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{- 18 \pm 8}{2 ( - 13 )}

将解分隔开

u_{1} = \frac{- 18 + 8}{2 ( - 13 )}, u_{2} = \frac{- 18 - 8}{2 ( - 13 )}

u = \frac{- 18 + 8}{2 ( - 13 )} : \frac{5}{13}

\frac{- 18 + 8}{2 ( - 13 )}

去除括号:

(- a) = - a

= \frac{- 18 + 8}{- 2 \cdot 13}

数字相加/相减:

- 18 + 8 = - 10

= \frac{- 10}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{- 10}{- 26}

使用分式法则:

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

= \frac{10}{26}

约分:

2

= \frac{5}{13}

u = \frac{- 18 - 8}{2 ( - 13 )} : 1

\frac{- 18 - 8}{2 ( - 13 )}

去除括号:

(- a) = - a

= \frac{- 18 - 8}{- 2 \cdot 13}

数字相减:

- 18 - 8 = - 26

= \frac{- 26}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{- 26}{- 26}

使用分式法则:

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

= \frac{26}{26}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = \frac{5}{13}, u = 1

u = cos (x)

代回

cos (x) = \frac{5}{13}, cos (x) = 1

cos (x) = \frac{5}{13}, cos (x) = 1

cos (x) = \frac{5}{13} : x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

cos (x) = \frac{5}{13}

使用反三角函数性质

cos (x) = \frac{5}{13}

cos (x) = \frac{5}{13}

的通解

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

x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 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

合并所有解

x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn, x = 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{5}{13}) + 2 πn

的解:

真

arccos (\frac{5}{13}) + 2 πn

代入

n = 1

arccos (\frac{5}{13}) + 2 π 1

对于

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

代入

x = arccos (\frac{5}{13}) + 2 π 1

2 sin (arccos (\frac{5}{13}) + 2 π 1) + 3 cos (arccos (\frac{5}{13}) + 2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

检验

2 π - arccos (\frac{5}{13}) + 2 πn

的解:

假

2 π - arccos (\frac{5}{13}) + 2 πn

代入

n = 1

2 π - arccos (\frac{5}{13}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{5}{13}) + 2 π 1

2 sin (2 π - arccos (\frac{5}{13}) + 2 π 1) + 3 cos (2 π - arccos (\frac{5}{13}) + 2 π 1) = 3

整理后得

- 0.69230 \dots = 3

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

2 sin (2 π 1) + 3 cos (2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

x = arccos (\frac{5}{13}) + 2 πn, x = 2 πn

以小数形式表示解

x = 1.17600 \dots + 2 πn, x = 2 πn

作图

流行的例子

cos (θ) = \frac{π}{3}

2 cos^{2} (x) = \frac{1}{2}

2sin(2x+(3pi)/2)=sqrt(3)

2 sin (2 x + \frac{3 π}{2}) = 3

2cos(θ)-2=6cos(θ)

2 cos (θ) - 2 = 6 cos (θ)

10sin^2(x)+15sin(x)+5=0

10 sin^{2} (x) + 15 sin (x) + 5 = 0