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受欢迎的三角函数 >

4sin(x)+3cos(x)=3

解答

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

解答

x = 1.85459 \dots + 2 πn, x = 2 πn

+1

度数

x = 106.26020 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3 cos (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

- 9 + 16 sin^{2} (x) + 18 cos (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 + 16 (1 - cos^{2} (x)) + 18 cos (x) - 9 cos^{2} (x)

化简

- 9 + 16 (1 - cos^{2} (x)) + 18 cos (x) - 9 cos^{2} (x) : 18 cos (x) - 25 cos^{2} (x) + 7

- 9 + 16 (1 - cos^{2} (x)) + 18 cos (x) - 9 cos^{2} (x)

乘开

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

数字相乘:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

= - 9 + 16 - 16 cos^{2} (x) + 18 cos (x) - 9 cos^{2} (x)

化简

- 9 + 16 - 16 cos^{2} (x) + 18 cos (x) - 9 cos^{2} (x) : 18 cos (x) - 25 cos^{2} (x) + 7

- 9 + 16 - 16 cos^{2} (x) + 18 cos (x) - 9 cos^{2} (x)

对同类项分组

= - 16 cos^{2} (x) + 18 cos (x) - 9 cos^{2} (x) - 9 + 16

同类项相加:

- 16 cos^{2} (x) - 9 cos^{2} (x) = - 25 cos^{2} (x)

= - 25 cos^{2} (x) + 18 cos (x) - 9 + 16

数字相加/相减:

- 9 + 16 = 7

= 18 cos (x) - 25 cos^{2} (x) + 7

= 18 cos (x) - 25 cos^{2} (x) + 7

= 18 cos (x) - 25 cos^{2} (x) + 7

7 + 18 cos (x) - 25 cos^{2} (x) = 0

用替代法求解

7 + 18 cos (x) - 25 cos^{2} (x) = 0

令

: cos (x) = u

7 + 18 u - 25 u^{2} = 0

7 + 18 u - 25 u^{2} = 0 : u = - \frac{7}{25}, u = 1

7 + 18 u - 25 u^{2} = 0

改写成标准形式

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

- 25 u^{2} + 18 u + 7 = 0

使用求根公式求解

- 25 u^{2} + 18 u + 7 = 0

二次方程求根公式:

若

a = - 25, b = 18, c = 7

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 25 ) \cdot 7}{2 ( - 25 )}

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 25 ) \cdot 7}{2 ( - 25 )}

1 8^{2} - 4 (- 25) \cdot 7 = 32

1 8^{2} - 4 (- 25) \cdot 7

使用法则

- (- a) = a

= 1 8^{2} + 4 \cdot 25 \cdot 7

数字相乘:

4 \cdot 25 \cdot 7 = 700

= 1 8^{2} + 700

1 8^{2} = 324

= 324 + 700

数字相加:

324 + 700 = 1024

= 1024

因式分解数字:

1024 = 3 2^{2}

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

3 2^{2} = 32

= 32

u_{1, 2} = \frac{- 18 \pm 32}{2 ( - 25 )}

将解分隔开

u_{1} = \frac{- 18 + 32}{2 ( - 25 )}, u_{2} = \frac{- 18 - 32}{2 ( - 25 )}

u = \frac{- 18 + 32}{2 ( - 25 )} : - \frac{7}{25}

\frac{- 18 + 32}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 18 + 32}{- 2 \cdot 25}

数字相加/相减:

- 18 + 32 = 14

= \frac{14}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{14}{- 50}

使用分式法则:

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

= - \frac{14}{50}

约分:

2

= - \frac{7}{25}

u = \frac{- 18 - 32}{2 ( - 25 )} : 1

\frac{- 18 - 32}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 18 - 32}{- 2 \cdot 25}

数字相减:

- 18 - 32 = - 50

= \frac{- 50}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{- 50}{- 50}

使用分式法则:

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

= \frac{50}{50}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - \frac{7}{25}, u = 1

u = cos (x)

代回

cos (x) = - \frac{7}{25}, cos (x) = 1

cos (x) = - \frac{7}{25}, cos (x) = 1

cos (x) = - \frac{7}{25} : x = arccos (- \frac{7}{25}) + 2 πn, x = - arccos (- \frac{7}{25}) + 2 πn

cos (x) = - \frac{7}{25}

使用反三角函数性质

cos (x) = - \frac{7}{25}

cos (x) = - \frac{7}{25}

的通解

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

x = arccos (- \frac{7}{25}) + 2 πn, x = - arccos (- \frac{7}{25}) + 2 πn

x = arccos (- \frac{7}{25}) + 2 πn, x = - arccos (- \frac{7}{25}) + 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{7}{25}) + 2 πn, x = - arccos (- \frac{7}{25}) + 2 πn, x = 2 πn

将解代入原方程进行验证

将它们代入

4 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{7}{25}) + 2 πn

的解:

真

arccos (- \frac{7}{25}) + 2 πn

代入

n = 1

arccos (- \frac{7}{25}) + 2 π 1

对于

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

代入

x = arccos (- \frac{7}{25}) + 2 π 1

4 sin (arccos (- \frac{7}{25}) + 2 π 1) + 3 cos (arccos (- \frac{7}{25}) + 2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

检验

- arccos (- \frac{7}{25}) + 2 πn

的解:

假

- arccos (- \frac{7}{25}) + 2 πn

代入

n = 1

- arccos (- \frac{7}{25}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{7}{25}) + 2 π 1

4 sin (- arccos (- \frac{7}{25}) + 2 π 1) + 3 cos (- arccos (- \frac{7}{25}) + 2 π 1) = 3

整理后得

- 4.68 = 3

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

4 sin (2 π 1) + 3 cos (2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

x = arccos (- \frac{7}{25}) + 2 πn, x = 2 πn

以小数形式表示解

x = 1.85459 \dots + 2 πn, x = 2 πn

作图

流行的例子

2 sin (θ) - 5 = 0

csc^5(x)-4csc(x)=0

csc^{5} (x) - 4 csc (x) = 0

arctan(x)=-pi/2

arctan (x) = - \frac{π}{2}

sec^2(x)=1+tan(x)

sec^{2} (x) = 1 + tan (x)

csc^2(θ)-2cot(θ)=0

csc^{2} (θ) - 2 cot (θ) = 0