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受欢迎的三角函数 >

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

解答

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

解答

x = - \frac{π + 4 πn}{4}, x = - \frac{π + 4 πn}{16}

+1

度数

x = - 4 5^{\circ} - 18 0^{\circ} n, x = - 11.2 5^{\circ} - 4 5^{\circ} n

求解步骤

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

两边减去

sin (5 x)

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

使用三角恒等式改写

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

利用以下特性:

- sin (x) = sin (- x)

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

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (\frac{π}{2} - 3 x) = sin (- (5 x))

sin (\frac{π}{2} - 3 x) = sin (- (5 x))

使用反三角函数性质

sin (\frac{π}{2} - 3 x) = sin (- (5 x))

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

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

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

- (5 x) = \frac{π}{2} - 3 x + 2 πn : x = - \frac{π + 4 πn}{4}

- (5 x) = \frac{π}{2} - 3 x + 2 πn

展开

- (5 x) : - 5 x

- (5 x)

去除括号:

(a) = a

= - 5 x

- 5 x = \frac{π}{2} - 3 x + 2 πn

将

3 x

para o lado esquerdo

- 5 x = \frac{π}{2} - 3 x + 2 πn

两边加上

3 x

- 5 x + 3 x = \frac{π}{2} - 3 x + 2 πn + 3 x

化简

- 2 x = \frac{π}{2} + 2 πn

- 2 x = \frac{π}{2} + 2 πn

两边除以

- 2

- 2 x = \frac{π}{2} + 2 πn

两边除以

- 2

\frac{- 2 x}{- 2} = \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

化简

\frac{- 2 x}{- 2} = \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

化简

\frac{- 2 x}{- 2} : x

\frac{- 2 x}{- 2}

使用分式法则:

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

= \frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2} : - \frac{π + 4 πn}{4}

\frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{\frac{π}{2} + 2 πn}{- 2}

使用分式法则:

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

= - \frac{\frac{π}{2} + 2 πn}{2}

化简

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + 2 πn \cdot 2}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= - \frac{\frac{π + 4 πn}{2}}{2}

化简

\frac{\frac{π + 4 πn}{2}}{2} : \frac{π + 4 πn}{4}

\frac{\frac{π + 4 πn}{2}}{2}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π + 4 πn}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{4}

= - \frac{π + 4 πn}{4}

x = - \frac{π + 4 πn}{4}

x = - \frac{π + 4 πn}{4}

x = - \frac{π + 4 πn}{4}

- (5 x) = π - (\frac{π}{2} - 3 x) + 2 πn : x = - \frac{π + 4 πn}{16}

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

展开

- (5 x) : - 5 x

- (5 x)

去除括号:

(a) = a

= - 5 x

展开

π - (\frac{π}{2} - 3 x) + 2 πn : π - \frac{π}{2} + 3 x + 2 πn

π - (\frac{π}{2} - 3 x) + 2 πn

- (\frac{π}{2} - 3 x) : - \frac{π}{2} + 3 x

- (\frac{π}{2} - 3 x)

打开括号

= - (\frac{π}{2}) - (- 3 x)

使用加减运算法则

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

= - \frac{π}{2} + 3 x

= π - \frac{π}{2} + 3 x + 2 πn

- 5 x = π - \frac{π}{2} + 3 x + 2 πn

将

3 x

para o lado esquerdo

- 5 x = π - \frac{π}{2} + 3 x + 2 πn

两边减去

3 x

- 5 x - 3 x = π - \frac{π}{2} + 3 x + 2 πn - 3 x

化简

- 8 x = π - \frac{π}{2} + 2 πn

- 8 x = π - \frac{π}{2} + 2 πn

两边除以

- 8

- 8 x = π - \frac{π}{2} + 2 πn

两边除以

- 8

\frac{- 8 x}{- 8} = \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8} + \frac{2 πn}{- 8}

化简

\frac{- 8 x}{- 8} = \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8} + \frac{2 πn}{- 8}

化简

\frac{- 8 x}{- 8} : x

\frac{- 8 x}{- 8}

使用分式法则:

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

= \frac{8 x}{8}

数字相除:

\frac{8}{8} = 1

= x

化简

\frac{π}{- 8} - \frac{\frac{π}{2}}{- 8} + \frac{2 πn}{- 8} : - \frac{π + 4 πn}{16}

\frac{π}{- 8} - \frac{\frac{π}{2}}{- 8} + \frac{2 πn}{- 8}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π - \frac{π}{2} + 2 πn}{- 8}

使用分式法则:

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

= - \frac{π - \frac{π}{2} + 2 πn}{8}

化简

π - \frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

π - \frac{π}{2} + 2 πn

将项转换为分式:

π = \frac{π 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{π 2}{2} - \frac{π}{2} + \frac{2 πn \cdot 2}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π 2 - π + 2 πn \cdot 2}{2}

π 2 - π + 2 πn \cdot 2 = π + 4 πn

π 2 - π + 2 πn \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= π + 4 πn

= \frac{π + 4 πn}{2}

= - \frac{\frac{π + 4 πn}{2}}{8}

化简

\frac{\frac{π + 4 πn}{2}}{8} : \frac{π + 4 πn}{16}

\frac{\frac{π + 4 πn}{2}}{8}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π + 4 πn}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{π + 4 πn}{16}

= - \frac{π + 4 πn}{16}

x = - \frac{π + 4 πn}{16}

x = - \frac{π + 4 πn}{16}

x = - \frac{π + 4 πn}{16}

x = - \frac{π + 4 πn}{4}, x = - \frac{π + 4 πn}{16}

x = - \frac{π + 4 πn}{4}, x = - \frac{π + 4 πn}{16}

作图

流行的例子

tan (θ) = 3.8

solvefor k,f=x^{2/3}+(10-x^2)^{0.5}sin(kpix)

so l v e f or k, f = x^{\frac{2}{3}} + (10 - x^{2})^{0.5} sin (kπ x)

sec(x)tan(x)=2sqrt(3)

sec (x) tan (x) = 23

cos (2 x + 6 0^{\circ}) = 0

cos (θ) = 0.6596