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

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 6 0^{\circ} + 24 0^{\circ} n, θ = 18 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{π}{3} + \frac{4 πn}{3}, θ = π + \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 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

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

化简

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

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

同类项相加:

2 θ + θ = 3 θ

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

同类项相加:

2 θ - θ = θ

= 2 cos (\frac{3 θ}{2}) cos (\frac{θ}{2})

= 2 cos (\frac{3 θ}{2}) cos (\frac{θ}{2})

2 cos (\frac{3 θ}{2}) cos (\frac{θ}{2}) = 0

分别求解每个部分

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

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

cos (\frac{3 θ}{2}) = 0

cos (\frac{3 θ}{2}) = 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}

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

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

解

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

\frac{3 θ}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{3 θ}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 \cdot 3 θ}{2} : 3 θ

\frac{2 \cdot 3 θ}{2}

数字相乘:

2 \cdot 3 = 6

= \frac{6 θ}{2}

数字相除:

\frac{6}{2} = 3

= 3 θ

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

3 θ = π + 4 πn

3 θ = π + 4 πn

3 θ = π + 4 πn

两边除以

3

3 θ = π + 4 πn

两边除以

3

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

化简

θ = \frac{π}{3} + \frac{4 πn}{3}

θ = \frac{π}{3} + \frac{4 πn}{3}

解

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

\frac{3 θ}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{3 θ}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 \cdot 3 θ}{2} : 3 θ

\frac{2 \cdot 3 θ}{2}

数字相乘:

2 \cdot 3 = 6

= \frac{6 θ}{2}

数字相除:

\frac{6}{2} = 3

= 3 θ

化简

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

3 θ = 3 π + 4 πn

3 θ = 3 π + 4 πn

3 θ = 3 π + 4 πn

两边除以

3

3 θ = 3 π + 4 πn

两边除以

3

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

化简

θ = π + \frac{4 πn}{3}

θ = π + \frac{4 πn}{3}

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

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

cos (\frac{θ}{2}) = 0

cos (\frac{θ}{2}) = 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}

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

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

解

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

\frac{θ}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

解

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

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

θ = 3 π + 4 πn

θ = 3 π + 4 πn

θ = 3 π + 4 πn

θ = π + 4 πn, θ = 3 π + 4 πn

合并所有解

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

合并重叠的区间

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

合并所有解

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

作图

流行的例子

2cos^2(x)+7sin(x)=5

2 cos^{2} (x) + 7 sin (x) = 5

tan^2(x)+sec(x)=1

tan^{2} (x) + sec (x) = 1

- 4 sin (2 x) = 0

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

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

sin (x) = 0.8