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sin(4x)+cos(2x)=0

解答

sin (4 x) + cos (2 x) = 0

解答

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}, x = \frac{7 π + 12 πn}{12}, x = \frac{11 π + 12 πn}{12}

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 10 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n

求解步骤

sin (4 x) + cos (2 x) = 0

令

: u = 2 x

sin (2 u) + cos (u) = 0

使用三角恒等式改写

cos (u) + sin (2 u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= cos (u) + 2 sin (u) cos (u)

cos (u) + 2 cos (u) sin (u) = 0

分解

cos (u) + 2 cos (u) sin (u) : cos (u) (2 sin (u) + 1)

cos (u) + 2 cos (u) sin (u)

因式分解出通项

cos (u)

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

cos (u) (2 sin (u) + 1) = 0

分别求解每个部分

cos (u) = 0 or 2 sin (u) + 1 = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

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}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

2 sin (u) + 1 = 0 : u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

2 sin (u) + 1 = 0

将

1

到右边

2 sin (u) + 1 = 0

两边减去

1

2 sin (u) + 1 - 1 = 0 - 1

化简

2 sin (u) = - 1

2 sin (u) = - 1

两边除以

2

2 sin (u) = - 1

两边除以

2

\frac{2 sin ( u )}{2} = \frac{- 1}{2}

化简

sin (u) = - \frac{1}{2}

sin (u) = - \frac{1}{2}

sin (u) = - \frac{1}{2}

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}

u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

u = 2 x

代回

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

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}

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}

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}

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}

x = \frac{π + 4 πn}{4}

x = \frac{π + 4 πn}{4}

x = \frac{π + 4 πn}{4}

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{4}

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π + 4 πn}{4}

\frac{\frac{3 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{3 π}{2} + 2 πn}{2}

化简

\frac{3 π}{2} + 2 πn : \frac{3 π + 4 πn}{2}

\frac{3 π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{3 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{3 π + 2 πn \cdot 2}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{2}

= \frac{\frac{3 π + 4 πn}{2}}{2}

使用分式法则:

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

= \frac{3 π + 4 πn}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

2 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π + 12 πn}{12}

2 x = \frac{7 π}{6} + 2 πn

两边除以

2

2 x = \frac{7 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π + 12 πn}{12}

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{\frac{7 π}{6} + 2 πn}{2}

化简

\frac{7 π}{6} + 2 πn : \frac{7 π + 12 πn}{6}

\frac{7 π}{6} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 6}{6}

= \frac{7 π}{6} + \frac{2 πn \cdot 6}{6}

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{7 π + 2 πn \cdot 6}{6}

数字相乘:

2 \cdot 6 = 12

= \frac{7 π + 12 πn}{6}

= \frac{\frac{7 π + 12 πn}{6}}{2}

使用分式法则:

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

= \frac{7 π + 12 πn}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{7 π + 12 πn}{12}

x = \frac{7 π + 12 πn}{12}

x = \frac{7 π + 12 πn}{12}

x = \frac{7 π + 12 πn}{12}

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π + 12 πn}{12}

2 x = \frac{11 π}{6} + 2 πn

两边除以

2

2 x = \frac{11 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π + 12 πn}{12}

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{\frac{11 π}{6} + 2 πn}{2}

化简

\frac{11 π}{6} + 2 πn : \frac{11 π + 12 πn}{6}

\frac{11 π}{6} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 6}{6}

= \frac{11 π}{6} + \frac{2 πn \cdot 6}{6}

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{11 π + 2 πn \cdot 6}{6}

数字相乘:

2 \cdot 6 = 12

= \frac{11 π + 12 πn}{6}

= \frac{\frac{11 π + 12 πn}{6}}{2}

使用分式法则:

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

= \frac{11 π + 12 πn}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{11 π + 12 πn}{12}

x = \frac{11 π + 12 πn}{12}

x = \frac{11 π + 12 πn}{12}

x = \frac{11 π + 12 πn}{12}

x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}, x = \frac{7 π + 12 πn}{12}, x = \frac{11 π + 12 πn}{12}

流行的例子

sin^2(θ)-11sin(θ)=0

sin^{2} (θ) - 11 sin (θ) = 0

sin (θ) = m

sin(θ)=(sqrt(3))/2 ,cos(θ)<0,0<θ<2pi

sin (θ) = \frac{3}{2}, cos (θ) < 0, 0 < θ < 2 π

tan(x)=1,0<= x<2pi

tan (x) = 1, 0 \leq x < 2 π

solvefor x,sin(2x)*cos(2x)=0.5 求解

x, sin (2 x) \cdot cos (2 x) = 0.5