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受欢迎的三角函数 >

sin(2θ)+sin(4θ)=0

解答

sin (2 θ) + sin (4 θ) = 0

解答

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

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 12 0^{\circ} n, θ = 6 0^{\circ} + 12 0^{\circ} n

求解步骤

sin (2 θ) + sin (4 θ) = 0

使用三角恒等式改写

sin (2 θ) + sin (4 θ)

使用和差化积恒等式:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 sin (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

化简

2 sin (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2}) : 2 cos (θ) sin (3 θ)

2 sin (\frac{2 θ + 4 θ}{2}) cos (\frac{2 θ - 4 θ}{2})

\frac{2 θ + 4 θ}{2} = 3 θ

\frac{2 θ + 4 θ}{2}

同类项相加:

2 θ + 4 θ = 6 θ

= \frac{6 θ}{2}

数字相除:

\frac{6}{2} = 3

= 3 θ

= 2 sin (3 θ) cos (\frac{2 θ - 4 θ}{2})

\frac{2 θ - 4 θ}{2} = - θ

\frac{2 θ - 4 θ}{2}

同类项相加:

2 θ - 4 θ = - 2 θ

= \frac{- 2 θ}{2}

使用分式法则:

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

= - \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= - θ

= 2 sin (3 θ) cos (- θ)

使用负角恒等式:

cos (- x) = cos (x)

= 2 cos (θ) sin (3 θ)

= 2 cos (θ) sin (3 θ)

2 cos (θ) sin (3 θ) = 0

分别求解每个部分

cos (θ) = 0 or sin (3 θ) = 0

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

cos (θ) = 0

cos (θ) = 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}

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

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

sin (3 θ) = 0 : θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

sin (3 θ) = 0

sin (3 θ) = 0

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 = 0 + 2 πn, 3 θ = π + 2 πn

3 θ = 0 + 2 πn, 3 θ = π + 2 πn

解

3 θ = 0 + 2 πn : θ = \frac{2 πn}{3}

3 θ = 0 + 2 πn

0 + 2 πn = 2 πn

3 θ = 2 πn

两边除以

3

3 θ = 2 πn

两边除以

3

\frac{3 θ}{3} = \frac{2 πn}{3}

化简

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

解

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

3 θ = π + 2 πn

两边除以

3

3 θ = π + 2 πn

两边除以

3

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

化简

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

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

合并所有解

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

作图

流行的例子

sqrt(3)tan(x)-1=0

3 tan (x) - 1 = 0

cos (4 x) = - \frac{1}{2}

16 cos^{2} (x) - 4 = 0

cos(2x)+cos(x)=-1

cos (2 x) + cos (x) = - 1

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

3 sin (x) = 2 cos^{2} (x)