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受欢迎的三角函数 >

solvefor x,sin(x)cos(2x)>0

解答

求解

x, sin (x) cos (2 x) > 0

解答

2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < π + 2 πn or - \frac{3 π}{4} + 2 πn < x < - \frac{π}{4} + 2 πn

求解步骤

sin (x) cos (2 x) > 0

利用以下特性:

cos (2 x) = 1 - 2 sin^{2} (x)

(1 - 2 sin^{2} (x)) sin (x) > 0

令

: u = sin (x)

(1 - 2 u^{2}) u > 0

(1 - 2 u^{2}) u > 0 : u < - \frac{2}{2} or 0 < u < \frac{2}{2}

(1 - 2 u^{2}) u > 0

分解

(1 - 2 u^{2}) u : - u (2 u + 1) (2 u - 1)

(1 - 2 u^{2}) u

分解

- 2 u^{2} + 1 : - (2 u + 1) (2 u - 1)

- 2 u^{2} + 1

因式分解出通项

- 1

= - (2 u^{2} - 1)

分解

2 u^{2} - 1 : (2 u + 1) (2 u - 1)

2 u^{2} - 1

将

2 u^{2} - 1

改写为

(2 u)^{2} - 1^{2}

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} u^{2} - 1

将

1

改写为

1^{2}

= (2)^{2} u^{2} - 1^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} u^{2} = (2 u)^{2}

= (2 u)^{2} - 1^{2}

= (2 u)^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 u)^{2} - 1^{2} = (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1)

= - (2 u + 1) (2 u - 1)

= - u (2 u + 1) (2 u - 1)

- u (2 u + 1) (2 u - 1) > 0

两边乘以

- 1

（改变不等式符号）

(- u (2 u + 1) (2 u - 1)) (- 1) < 0 \cdot (- 1)

化简

u (2 u + 1) (2 u - 1) < 0

确定区间

确定

u (2 u + 1) (2 u - 1)

符号

确定

u

符号

u = 0

u < 0

u > 0

确定

2 u + 1

符号

2 u + 1 = 0 : u = - \frac{2}{2}

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

2 u + 1 < 0 : u < - \frac{2}{2}

2 u + 1 < 0

将

1

到右边

2 u + 1 < 0

两边减去

1

2 u + 1 - 1 < 0 - 1

化简

2 u < - 1

2 u < - 1

两边除以

2

2 u < - 1

两边除以

2

\frac{2 u}{2} < \frac{- 1}{2}

化简

\frac{2 u}{2} < \frac{- 1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u < - \frac{2}{2}

u < - \frac{2}{2}

u < - \frac{2}{2}

2 u + 1 > 0 : u > - \frac{2}{2}

2 u + 1 > 0

将

1

到右边

2 u + 1 > 0

两边减去

1

2 u + 1 - 1 > 0 - 1

化简

2 u > - 1

2 u > - 1

两边除以

2

2 u > - 1

两边除以

2

\frac{2 u}{2} > \frac{- 1}{2}

化简

\frac{2 u}{2} > \frac{- 1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u > - \frac{2}{2}

u > - \frac{2}{2}

u > - \frac{2}{2}

确定

2 u - 1

符号

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

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

2 u - 1 < 0 : u < \frac{2}{2}

2 u - 1 < 0

将

1

到右边

2 u - 1 < 0

两边加上

1

2 u - 1 + 1 < 0 + 1

化简

2 u < 1

2 u < 1

两边除以

2

2 u < 1

两边除以

2

\frac{2 u}{2} < \frac{1}{2}

化简

\frac{2 u}{2} < \frac{1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

u < \frac{2}{2}

u < \frac{2}{2}

u < \frac{2}{2}

2 u - 1 > 0 : u > \frac{2}{2}

2 u - 1 > 0

将

1

到右边

2 u - 1 > 0

两边加上

1

2 u - 1 + 1 > 0 + 1

化简

2 u > 1

2 u > 1

两边除以

2

2 u > 1

两边除以

2

\frac{2 u}{2} > \frac{1}{2}

化简

\frac{2 u}{2} > \frac{1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

u > \frac{2}{2}

u > \frac{2}{2}

u > \frac{2}{2}

总结如下表：

u 2 u + 1 2 u - 1 u (2 u + 1) (2 u - 1) u < - \frac{2}{2} - - - - u = - \frac{2}{2} - 0 - 0 - \frac{2}{2} < u < 0 - + - + u = 0 0 + - 0 0 < u < \frac{2}{2} + + - - u = \frac{2}{2} + + 00 u > \frac{2}{2} + + + +

确定满足所需条件的区间：

< 0

u < - \frac{2}{2} or 0 < u < \frac{2}{2}

u < - \frac{2}{2} or 0 < u < \frac{2}{2}

u < - \frac{2}{2} or 0 < u < \frac{2}{2}

u = sin (x)

代回

sin (x) < - \frac{2}{2} or 0 < sin (x) < \frac{2}{2}

sin (x) < - \frac{2}{2} : - \frac{3 π}{4} + 2 πn < x < - \frac{π}{4} + 2 πn

sin (x) < - \frac{2}{2}

对于

sin (x) < a

，若

- 1 < a \leq 1

，则

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (- \frac{2}{2}) + 2 πn < x < arcsin (- \frac{2}{2}) + 2 πn

化简

- π - arcsin (- \frac{2}{2}) : - \frac{3 π}{4}

- π - arcsin (- \frac{2}{2})

arcsin (- \frac{2}{2}) = - \frac{π}{4}

arcsin (- \frac{2}{2})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2}{2}) = - arcsin (\frac{2}{2})

= - arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

= - π - (- \frac{π}{4})

化简

- π - (- \frac{π}{4})

使用法则

- (- a) = a

= - π + \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}

= - \frac{π 4}{4} + \frac{π}{4}

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{- π 4 + π}{4}

同类项相加:

- 4 π + π = - 3 π

= \frac{- 3 π}{4}

使用分式法则:

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

= - \frac{3 π}{4}

= - \frac{3 π}{4}

化简

arcsin (- \frac{2}{2}) : - \frac{π}{4}

arcsin (- \frac{2}{2})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2}{2}) = - arcsin (\frac{2}{2})

= - arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

- \frac{3 π}{4} + 2 πn < x < - \frac{π}{4} + 2 πn

0 < sin (x) < \frac{2}{2} : 2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < π + 2 πn

0 < sin (x) < \frac{2}{2}

若

a < u < b

，则

a < u and u < b

0 < sin (x) and sin (x) < \frac{2}{2}

0 < sin (x) : 2 πn < x < π + 2 πn

0 < sin (x)

交换两边

sin (x) > 0

对于

sin (x) > a

，若

- 1 \leq a < 1

，则

arcsin (a) + 2 πn < x < π - arcsin (a) + 2 πn

arcsin (0) + 2 πn < x < π - arcsin (0) + 2 πn

化简

arcsin (0) : 0

arcsin (0)

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

化简

π - arcsin (0) : π

π - arcsin (0)

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - 0

π - 0 = π

= π

0 + 2 πn < x < π + 2 πn

化简

2 πn < x < π + 2 πn

sin (x) < \frac{2}{2} : - \frac{5 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

sin (x) < \frac{2}{2}

对于

sin (x) < a

，若

- 1 < a \leq 1

，则

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (\frac{2}{2}) + 2 πn < x < arcsin (\frac{2}{2}) + 2 πn

化简

- π - arcsin (\frac{2}{2}) : - \frac{5 π}{4}

- π - arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{4}

化简

- π - \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}

= - \frac{π 4}{4} - \frac{π}{4}

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{- π 4 - π}{4}

同类项相加:

- 4 π - π = - 5 π

= \frac{- 5 π}{4}

使用分式法则:

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

= - \frac{5 π}{4}

= - \frac{5 π}{4}

化简

arcsin (\frac{2}{2}) : \frac{π}{4}

arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

- \frac{5 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

合并区间

2 πn < x < π + 2 πn and - \frac{5 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

合并重叠的区间

2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < π + 2 πn

合并区间

- \frac{3 π}{4} + 2 πn < x < - \frac{π}{4} + 2 πn or (2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < π + 2 πn)

合并重叠的区间

2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < π + 2 πn or - \frac{3 π}{4} + 2 πn < x < - \frac{π}{4} + 2 πn

流行的例子

cos (2 x) \geq 1

cos(x)-sin(x)>=-1

cos (x) - sin (x) \geq - 1

sin(x)-sqrt(3)cos(x)>0

sin (x) - 3 cos (x) > 0

sin^{2} (x) > 1

pi/2-arctan(x^4)<0.0001

\frac{π}{2} - arctan (x^{4}) < 0.0001