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受欢迎的三角函数 >

2sin^2(3θ)-1=0

解答

2 sin^{2} (3 θ) - 1 = 0

解答

θ = \frac{π}{12} + \frac{2 πn}{3}, θ = \frac{π}{4} + \frac{2 πn}{3}, θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \frac{2 πn}{3}

+1

度数

θ = 1 5^{\circ} + 12 0^{\circ} n, θ = 4 5^{\circ} + 12 0^{\circ} n, θ = 7 5^{\circ} + 12 0^{\circ} n, θ = 10 5^{\circ} + 12 0^{\circ} n

求解步骤

2 sin^{2} (3 θ) - 1 = 0

用替代法求解

2 sin^{2} (3 θ) - 1 = 0

令

: sin (3 θ) = u

2 u^{2} - 1 = 0

2 u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{2} - 1 = 0

将

1

到右边

2 u^{2} - 1 = 0

两边加上

1

2 u^{2} - 1 + 1 = 0 + 1

化简

2 u^{2} = 1

2 u^{2} = 1

两边除以

2

2 u^{2} = 1

两边除以

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

化简

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

u = sin (3 θ)

代回

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

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

sin (3 θ) = \frac{1}{2} : θ = \frac{π}{12} + \frac{2 πn}{3}, θ = \frac{π}{4} + \frac{2 πn}{3}

sin (3 θ) = \frac{1}{2}

sin (3 θ) = \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 θ = \frac{π}{4} + 2 πn, 3 θ = \frac{3 π}{4} + 2 πn

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

解

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

3 θ = \frac{π}{4} + 2 πn

两边除以

3

3 θ = \frac{π}{4} + 2 πn

两边除以

3

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

化简

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

化简

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

数字相除:

\frac{3}{3} = 1

= θ

化简

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

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

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

使用分式法则:

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

= \frac{π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{2 πn}{3}

θ = \frac{π}{12} + \frac{2 πn}{3}

θ = \frac{π}{12} + \frac{2 πn}{3}

θ = \frac{π}{12} + \frac{2 πn}{3}

解

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

3 θ = \frac{3 π}{4} + 2 πn

两边除以

3

3 θ = \frac{3 π}{4} + 2 πn

两边除以

3

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

化简

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

化简

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

数字相除:

\frac{3}{3} = 1

= θ

化简

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

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

\frac{\frac{3 π}{4}}{3} = \frac{π}{4}

\frac{\frac{3 π}{4}}{3}

使用分式法则:

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

= \frac{3 π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{3 π}{12}

约分:

3

= \frac{π}{4}

= \frac{π}{4} + \frac{2 πn}{3}

θ = \frac{π}{4} + \frac{2 πn}{3}

θ = \frac{π}{4} + \frac{2 πn}{3}

θ = \frac{π}{4} + \frac{2 πn}{3}

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

sin (3 θ) = - \frac{1}{2} : θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \frac{2 πn}{3}

sin (3 θ) = - \frac{1}{2}

sin (3 θ) = - \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 θ = \frac{5 π}{4} + 2 πn, 3 θ = \frac{7 π}{4} + 2 πn

3 θ = \frac{5 π}{4} + 2 πn, 3 θ = \frac{7 π}{4} + 2 πn

解

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

3 θ = \frac{5 π}{4} + 2 πn

两边除以

3

3 θ = \frac{5 π}{4} + 2 πn

两边除以

3

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

化简

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

化简

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

数字相除:

\frac{3}{3} = 1

= θ

化简

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

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

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

\frac{\frac{5 π}{4}}{3}

使用分式法则:

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

= \frac{5 π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{2 πn}{3}

θ = \frac{5 π}{12} + \frac{2 πn}{3}

θ = \frac{5 π}{12} + \frac{2 πn}{3}

θ = \frac{5 π}{12} + \frac{2 πn}{3}

解

3 θ = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{12} + \frac{2 πn}{3}

3 θ = \frac{7 π}{4} + 2 πn

两边除以

3

3 θ = \frac{7 π}{4} + 2 πn

两边除以

3

\frac{3 θ}{3} = \frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

化简

\frac{3 θ}{3} = \frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

化简

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

数字相除:

\frac{3}{3} = 1

= θ

化简

\frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3} : \frac{7 π}{12} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} = \frac{7 π}{12}

\frac{\frac{7 π}{4}}{3}

使用分式法则:

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

= \frac{7 π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{7 π}{12}

= \frac{7 π}{12} + \frac{2 πn}{3}

θ = \frac{7 π}{12} + \frac{2 πn}{3}

θ = \frac{7 π}{12} + \frac{2 πn}{3}

θ = \frac{7 π}{12} + \frac{2 πn}{3}

θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \frac{2 πn}{3}

合并所有解

θ = \frac{π}{12} + \frac{2 πn}{3}, θ = \frac{π}{4} + \frac{2 πn}{3}, θ = \frac{5 π}{12} + \frac{2 πn}{3}, θ = \frac{7 π}{12} + \frac{2 πn}{3}

作图

流行的例子

4sin(4θ-pi/3)+3=5

4 sin (4 θ - \frac{π}{3}) + 3 = 5

2 tan (x) - 3 = 0

-2cos(2x-pi/3)=2sin(2x-pi/6)

- 2 cos (2 x - \frac{π}{3}) = 2 sin (2 x - \frac{π}{6})

tan(x)=5sin(pi/4)sin((3pi)/4)

tan (x) = 5 sin (\frac{π}{4}) sin (\frac{3 π}{4})

cot (θ) = - \frac{12}{5}