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受欢迎的三角函数 >

2sin(3x)+1=0

解答

2 sin (3 x) + 1 = 0

解答

x = \frac{7 π}{18} + \frac{2 πn}{3}, x = \frac{11 π}{18} + \frac{2 πn}{3}

+1

度数

x = 7 0^{\circ} + 12 0^{\circ} n, x = 11 0^{\circ} + 12 0^{\circ} n

求解步骤

2 sin (3 x) + 1 = 0

将

1

到右边

2 sin (3 x) + 1 = 0

两边减去

1

2 sin (3 x) + 1 - 1 = 0 - 1

化简

2 sin (3 x) = - 1

2 sin (3 x) = - 1

两边除以

2

2 sin (3 x) = - 1

两边除以

2

\frac{2 sin ( 3 x )}{2} = \frac{- 1}{2}

化简

sin (3 x) = - \frac{1}{2}

sin (3 x) = - \frac{1}{2}

sin (3 x) = - \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}

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

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

解

3 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{18} + \frac{2 πn}{3}

3 x = \frac{7 π}{6} + 2 πn

两边除以

3

3 x = \frac{7 π}{6} + 2 πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{7 π}{6}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{7 π}{6}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{7 π}{6}}{3} + \frac{2 πn}{3} : \frac{7 π}{18} + \frac{2 πn}{3}

\frac{\frac{7 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{7 π}{6}}{3} = \frac{7 π}{18}

\frac{\frac{7 π}{6}}{3}

使用分式法则:

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

= \frac{7 π}{6 \cdot 3}

数字相乘:

6 \cdot 3 = 18

= \frac{7 π}{18}

= \frac{7 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}

解

3 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{18} + \frac{2 πn}{3}

3 x = \frac{11 π}{6} + 2 πn

两边除以

3

3 x = \frac{11 π}{6} + 2 πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3} : \frac{11 π}{18} + \frac{2 πn}{3}

\frac{\frac{11 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{11 π}{6}}{3} = \frac{11 π}{18}

\frac{\frac{11 π}{6}}{3}

使用分式法则:

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

= \frac{11 π}{6 \cdot 3}

数字相乘:

6 \cdot 3 = 18

= \frac{11 π}{18}

= \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{11 π}{18} + \frac{2 πn}{3}

x = \frac{7 π}{18} + \frac{2 πn}{3}, x = \frac{11 π}{18} + \frac{2 πn}{3}

作图

流行的例子

sqrt(3)sin(θ)=cos(θ),0<= θ<= 2pi

3 sin (θ) = cos (θ), 0 \leq θ \leq 2 π

cos(2θ)=-(sqrt(2))/2

cos (2 θ) = - \frac{2}{2}

tan(θ/2+pi/3)=-1

tan (\frac{θ}{2} + \frac{π}{3}) = - 1

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

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

sec(x)=-sqrt(2),0<= x<= 2pi

sec (x) = - 2, 0 \leq x \leq 2 π