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受欢迎的三角函数 >

3sin(x)+5cos(x)+5=0

解答

3 sin (x) + 5 cos (x) + 5 = 0

解答

x = π + 2 πn, x = π + 1.08083 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 241.92751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (x) + 5 cos (x) + 5 = 0

两边减去

5 cos (x)

3 sin (x) + 5 = - 5 cos (x)

两边进行平方

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

两边减去

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

(3 sin (x) + 5)^{2} - 25 cos^{2} (x) = 0

使用三角恒等式改写

(5 + 3 sin (x))^{2} - 25 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= (5 + 3 sin (x))^{2} - 25 (1 - sin^{2} (x))

化简

(5 + 3 sin (x))^{2} - 25 (1 - sin^{2} (x)) : 34 sin^{2} (x) + 30 sin (x)

(5 + 3 sin (x))^{2} - 25 (1 - sin^{2} (x))

(5 + 3 sin (x))^{2} : 25 + 30 sin (x) + 9 sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 5, b = 3 sin (x)

= 5^{2} + 2 \cdot 5 \cdot 3 sin (x) + (3 sin (x))^{2}

化简

5^{2} + 2 \cdot 5 \cdot 3 sin (x) + (3 sin (x))^{2} : 25 + 30 sin (x) + 9 sin^{2} (x)

5^{2} + 2 \cdot 5 \cdot 3 sin (x) + (3 sin (x))^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

2 \cdot 5 \cdot 3 sin (x) = 30 sin (x)

2 \cdot 5 \cdot 3 sin (x)

数字相乘:

2 \cdot 5 \cdot 3 = 30

= 30 sin (x)

(3 sin (x))^{2} = 9 sin^{2} (x)

(3 sin (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

3^{2} = 9

= 9 sin^{2} (x)

= 25 + 30 sin (x) + 9 sin^{2} (x)

= 25 + 30 sin (x) + 9 sin^{2} (x)

= 25 + 30 sin (x) + 9 sin^{2} (x) - 25 (1 - sin^{2} (x))

乘开

- 25 (1 - sin^{2} (x)) : - 25 + 25 sin^{2} (x)

- 25 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 25, b = 1, c = sin^{2} (x)

= - 25 \cdot 1 - (- 25) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 25 \cdot 1 + 25 sin^{2} (x)

数字相乘:

25 \cdot 1 = 25

= - 25 + 25 sin^{2} (x)

= 25 + 30 sin (x) + 9 sin^{2} (x) - 25 + 25 sin^{2} (x)

化简

25 + 30 sin (x) + 9 sin^{2} (x) - 25 + 25 sin^{2} (x) : 34 sin^{2} (x) + 30 sin (x)

25 + 30 sin (x) + 9 sin^{2} (x) - 25 + 25 sin^{2} (x)

对同类项分组

= 30 sin (x) + 9 sin^{2} (x) + 25 sin^{2} (x) + 25 - 25

同类项相加:

9 sin^{2} (x) + 25 sin^{2} (x) = 34 sin^{2} (x)

= 30 sin (x) + 34 sin^{2} (x) + 25 - 25

25 - 25 = 0

= 34 sin^{2} (x) + 30 sin (x)

= 34 sin^{2} (x) + 30 sin (x)

= 34 sin^{2} (x) + 30 sin (x)

30 sin (x) + 34 sin^{2} (x) = 0

用替代法求解

30 sin (x) + 34 sin^{2} (x) = 0

令

: sin (x) = u

30 u + 34 u^{2} = 0

30 u + 34 u^{2} = 0 : u = 0, u = - \frac{15}{17}

30 u + 34 u^{2} = 0

改写成标准形式

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

34 u^{2} + 30 u = 0

使用求根公式求解

34 u^{2} + 30 u = 0

二次方程求根公式:

若

a = 34, b = 30, c = 0

u_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 34 \cdot 0}{2 \cdot 34}

u_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 34 \cdot 0}{2 \cdot 34}

3 0^{2} - 4 \cdot 34 \cdot 0 = 30

3 0^{2} - 4 \cdot 34 \cdot 0

使用法则

0 \cdot a = 0

= 3 0^{2} - 0

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

= 3 0^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 30

u_{1, 2} = \frac{- 30 \pm 30}{2 \cdot 34}

将解分隔开

u_{1} = \frac{- 30 + 30}{2 \cdot 34}, u_{2} = \frac{- 30 - 30}{2 \cdot 34}

u = \frac{- 30 + 30}{2 \cdot 34} : 0

\frac{- 30 + 30}{2 \cdot 34}

数字相加/相减:

- 30 + 30 = 0

= \frac{0}{2 \cdot 34}

数字相乘:

2 \cdot 34 = 68

= \frac{0}{68}

使用法则

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

= 0

u = \frac{- 30 - 30}{2 \cdot 34} : - \frac{15}{17}

\frac{- 30 - 30}{2 \cdot 34}

数字相减:

- 30 - 30 = - 60

= \frac{- 60}{2 \cdot 34}

数字相乘:

2 \cdot 34 = 68

= \frac{- 60}{68}

使用分式法则:

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

= - \frac{60}{68}

约分:

4

= - \frac{15}{17}

二次方程组的解是:

u = 0, u = - \frac{15}{17}

u = sin (x)

代回

sin (x) = 0, sin (x) = - \frac{15}{17}

sin (x) = 0, sin (x) = - \frac{15}{17}

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

sin (x) = - \frac{15}{17} : x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

sin (x) = - \frac{15}{17}

使用反三角函数性质

sin (x) = - \frac{15}{17}

sin (x) = - \frac{15}{17}

的通解

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

x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

将解代入原方程进行验证

将它们代入

3 sin (x) + 5 cos (x) + 5 = 0

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

假

2 πn

代入

n = 1

2 π 1

对于

3 sin (x) + 5 cos (x) + 5 = 0

代入

x = 2 π 1

3 sin (2 π 1) + 5 cos (2 π 1) + 5 = 0

整理后得

10 = 0

\Rightarrow 假

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

3 sin (x) + 5 cos (x) + 5 = 0

代入

x = π + 2 π 1

3 sin (π + 2 π 1) + 5 cos (π + 2 π 1) + 5 = 0

整理后得

0 = 0

\Rightarrow 真

检验

arcsin (- \frac{15}{17}) + 2 πn

的解:

假

arcsin (- \frac{15}{17}) + 2 πn

代入

n = 1

arcsin (- \frac{15}{17}) + 2 π 1

对于

3 sin (x) + 5 cos (x) + 5 = 0

代入

x = arcsin (- \frac{15}{17}) + 2 π 1

3 sin (arcsin (- \frac{15}{17}) + 2 π 1) + 5 cos (arcsin (- \frac{15}{17}) + 2 π 1) + 5 = 0

整理后得

4.70588 \dots = 0

\Rightarrow 假

检验

π + arcsin (\frac{15}{17}) + 2 πn

的解:

真

π + arcsin (\frac{15}{17}) + 2 πn

代入

n = 1

π + arcsin (\frac{15}{17}) + 2 π 1

对于

3 sin (x) + 5 cos (x) + 5 = 0

代入

x = π + arcsin (\frac{15}{17}) + 2 π 1

3 sin (π + arcsin (\frac{15}{17}) + 2 π 1) + 5 cos (π + arcsin (\frac{15}{17}) + 2 π 1) + 5 = 0

整理后得

0 = 0

\Rightarrow 真

x = π + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

以小数形式表示解

x = π + 2 πn, x = π + 1.08083 \dots + 2 πn

作图

流行的例子

4 cos^{2} (x) + 2 = 3

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

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

sin(θ)sec(θ)-sin(θ)=0

sin (θ) sec (θ) - sin (θ) = 0

r = 2 a cos (x)

tan(t)=-1/(sqrt(3)),-pi<t<= pi

tan (t) = - \frac{1}{3}, - π < t \leq π