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受欢迎的三角函数 >

sin(x+pi/3)+sin(x-pi/3)=1,0<= x<= 2pi

解答

sin (x + \frac{π}{3}) + sin (x - \frac{π}{3}) = 1, 0 \leq x \leq 2 π

解答

x = \frac{π}{2}

+1

度数

x = 9 0^{\circ}

求解步骤

sin (x + \frac{π}{3}) + sin (x - \frac{π}{3}) = 1, 0 \leq x \leq 2 π

使用三角恒等式改写

sin (x + \frac{π}{3}) + sin (x - \frac{π}{3})

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{x + \frac{π}{3} + x - \frac{π}{3}}{2}) cos (\frac{x + \frac{π}{3} - ( x - \frac{π}{3} )}{2})

化简

2 sin (\frac{x + \frac{π}{3} + x - \frac{π}{3}}{2}) cos (\frac{x + \frac{π}{3} - ( x - \frac{π}{3} )}{2}) : sin (x)

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

\frac{x + \frac{π}{3} + x - \frac{π}{3}}{2} = x

\frac{x + \frac{π}{3} + x - \frac{π}{3}}{2}

x + \frac{π}{3} + x - \frac{π}{3} = 2 x

x + \frac{π}{3} + x - \frac{π}{3}

对同类项分组

= x + x + \frac{π}{3} - \frac{π}{3}

同类项相加:

x + x = 2 x

= 2 x + \frac{π}{3} - \frac{π}{3}

同类项相加:

\frac{π}{3} - \frac{π}{3} = 0

= 2 x

= \frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

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

\frac{x + \frac{π}{3} - ( x - \frac{π}{3} )}{2} = \frac{π}{3}

\frac{x + \frac{π}{3} - ( x - \frac{π}{3} )}{2}

化简

x + \frac{π}{3} - (x - \frac{π}{3}) : \frac{2 π}{3}

x + \frac{π}{3} - (x - \frac{π}{3})

将项转换为分式:

x = \frac{x 3}{3}, (x - \frac{π}{3}) = \frac{( x - \frac{π}{3} ) 3}{3}

= \frac{x \cdot 3}{3} + \frac{π}{3} - \frac{( x - \frac{π}{3} ) \cdot 3}{3}

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{x \cdot 3 + π - ( x - \frac{π}{3} ) \cdot 3}{3}

乘开

x \cdot 3 + π - (x - \frac{π}{3}) \cdot 3 : 2 π

x \cdot 3 + π - (x - \frac{π}{3}) \cdot 3

= 3 x + π - 3 (x - \frac{π}{3})

乘开

- 3 (x - \frac{π}{3}) : - 3 x + π

- 3 (x - \frac{π}{3})

使用分配律:

a (b - c) = ab - a c

a = - 3, b = x, c = \frac{π}{3}

= - 3 x - (- 3) \frac{π}{3}

使用加减运算法则

- (- a) = a

= - 3 x + 3 \cdot \frac{π}{3}

3 \cdot \frac{π}{3} = π

3 \cdot \frac{π}{3}

分式相乘:

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

= \frac{π 3}{3}

约分:

3

= π

= - 3 x + π

= x \cdot 3 + π - 3 x + π

化简

x \cdot 3 + π - 3 x + π : 2 π

x \cdot 3 + π - 3 x + π

对同类项分组

= 3 x - 3 x + π + π

同类项相加:

3 x - 3 x = 0

= π + π

同类项相加:

π + π = 2 π

= 2 π

= 2 π

= \frac{2 π}{3}

= \frac{\frac{2 π}{3}}{2}

使用分式法则:

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

= \frac{2 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π}{6}

约分:

2

= \frac{π}{3}

= 2 cos (\frac{π}{3}) sin (x)

化简

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

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{1}{2}

= 2 \cdot \frac{1}{2} sin (x)

分式相乘:

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

= \frac{1 \cdot 2}{2} sin (x)

约分:

2

= sin (x) \cdot 1

乘以:

sin (x) \cdot 1 = sin (x)

= sin (x)

= sin (x)

sin (x) = 1

sin (x) = 1

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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{2}

作图

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