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受欢迎的三角函数 >

-tan(2θ)+2sin(θ)=0

解答

- tan (2 θ) + 2 sin (θ) = 0

解答

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

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 24 0^{\circ} n, θ = 12 0^{\circ} + 24 0^{\circ} n

求解步骤

- tan (2 θ) + 2 sin (θ) = 0

用 sin, cos 表示

- tan (2 θ) + 2 sin (θ)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

化简

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

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

将项转换为分式:

2 sin (θ) = \frac{2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

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

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{- sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

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

\frac{- sin ( 2 θ ) + 2 cos ( 2 θ ) sin ( θ )}{cos ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin (2 θ) + 2 cos (2 θ) sin (θ) = 0

使用三角恒等式改写

- sin (2 θ) + 2 cos (2 θ) sin (θ)

使用倍角公式:

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

= - 2 sin (θ) cos (θ) + 2 cos (2 θ) sin (θ)

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) = 0

分解

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) : 2 sin (θ) (cos (2 θ) - cos (θ))

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ)

改写为

= 2 sin (θ) cos (2 θ) - 2 sin (θ) cos (θ)

因式分解出通项

2 sin (θ)

= 2 sin (θ) (cos (2 θ) - cos (θ))

2 sin (θ) (cos (2 θ) - cos (θ)) = 0

分别求解每个部分

sin (θ) = 0 or cos (2 θ) - cos (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (2 θ) - cos (θ) = 0 : θ = \frac{4 πn}{3}, θ = \frac{2 π}{3} + \frac{4 πn}{3}

cos (2 θ) - cos (θ) = 0

使用三角恒等式改写

cos (2 θ) - cos (θ)

使用和差化积恒等式:

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

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

化简

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

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

同类项相加:

2 θ + θ = 3 θ

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

同类项相加:

2 θ - θ = θ

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

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

- 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2}) = 0

分别求解每个部分

sin (\frac{3 θ}{2}) = 0 or sin (\frac{θ}{2}) = 0

sin (\frac{3 θ}{2}) = 0 : θ = \frac{4 πn}{3}, θ = \frac{2 π}{3} + \frac{4 πn}{3}

sin (\frac{3 θ}{2}) = 0

sin (\frac{3 θ}{2}) = 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}

\frac{3 θ}{2} = 0 + 2 πn, \frac{3 θ}{2} = π + 2 πn

\frac{3 θ}{2} = 0 + 2 πn, \frac{3 θ}{2} = π + 2 πn

解

\frac{3 θ}{2} = 0 + 2 πn : θ = \frac{4 πn}{3}

\frac{3 θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{3 θ}{2} = 2 πn

在两边乘以

2

\frac{3 θ}{2} = 2 πn

在两边乘以

2

\frac{2 \cdot 3 θ}{2} = 2 \cdot 2 πn

化简

3 θ = 4 πn

3 θ = 4 πn

两边除以

3

3 θ = 4 πn

两边除以

3

\frac{3 θ}{3} = \frac{4 πn}{3}

化简

θ = \frac{4 πn}{3}

θ = \frac{4 πn}{3}

解

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

\frac{3 θ}{2} = π + 2 πn

在两边乘以

2

\frac{3 θ}{2} = π + 2 πn

在两边乘以

2

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

化简

3 θ = 2 π + 4 πn

3 θ = 2 π + 4 πn

两边除以

3

3 θ = 2 π + 4 πn

两边除以

3

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

化简

θ = \frac{2 π}{3} + \frac{4 πn}{3}

θ = \frac{2 π}{3} + \frac{4 πn}{3}

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

sin (\frac{θ}{2}) = 0 : θ = 4 πn, θ = 2 π + 4 πn

sin (\frac{θ}{2}) = 0

sin (\frac{θ}{2}) = 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}

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

解

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot 2 πn

化简

θ = 4 πn

θ = 4 πn

解

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

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

化简

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

合并所有解

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

合并重叠的区间

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

合并所有解

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

作图

流行的例子

-3cos^2(θ)+6=sin(θ)+5

- 3 cos^{2} (θ) + 6 = sin (θ) + 5

cos^2(θ)+4cos(θ)=0

cos^{2} (θ) + 4 cos (θ) = 0

2 sin (x) = 2

tan(θ)=(-1)/(sqrt(3))

tan (θ) = \frac{- 1}{3}

cos(x)=-1/2 ,0<= x<= 2pi

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π