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受欢迎的三角函数 >

sin(4θ)=-(sqrt(3))/2

解答

sin (4 θ) = - \frac{3}{2}

解答

θ = \frac{π}{3} + \frac{πn}{2}, θ = \frac{5 π}{12} + \frac{πn}{2}

+1

度数

θ = 6 0^{\circ} + 9 0^{\circ} n, θ = 7 5^{\circ} + 9 0^{\circ} n

求解步骤

sin (4 θ) = - \frac{3}{2}

sin (4 θ) = - \frac{3}{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}

4 θ = \frac{4 π}{3} + 2 πn, 4 θ = \frac{5 π}{3} + 2 πn

4 θ = \frac{4 π}{3} + 2 πn, 4 θ = \frac{5 π}{3} + 2 πn

解

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

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

\frac{4 θ}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 θ}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

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

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

\frac{\frac{4 π}{3}}{4} = \frac{π}{3}

\frac{\frac{4 π}{3}}{4}

使用分式法则:

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

= \frac{4 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{4 π}{12}

约分:

4

= \frac{π}{3}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

解

4 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{12} + \frac{πn}{2}

4 θ = \frac{5 π}{3} + 2 πn

两边除以

4

4 θ = \frac{5 π}{3} + 2 πn

两边除以

4

\frac{4 θ}{4} = \frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 θ}{4} = \frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{5 π}{3}}{4} = \frac{5 π}{12}

\frac{\frac{5 π}{3}}{4}

使用分式法则:

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

= \frac{5 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{5 π}{12}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}, θ = \frac{5 π}{12} + \frac{πn}{2}

作图

流行的例子

1 = 6 cos (x)

solvefor x,sec(x)=0

so l v e f or x, sec (x) = 0

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

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

2cos(θ)sin(θ)-cos(θ)=0

2 cos (θ) sin (θ) - cos (θ) = 0

-sin(x)=cos(2x)

- sin (x) = cos (2 x)