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受欢迎的三角函数 >

2sin(2x)+1/2 >= 0

解答

2 sin (2 x) + \frac{1}{2} \geq 0

解答

- \frac{arcsin ( \frac{1}{4} )}{2} + πn \leq x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

+2

间隔符号

[- \frac{arcsin ( \frac{1}{4} )}{2} + πn, \frac{π + arcsin ( \frac{1}{4} )}{2} + πn]

十进制

- 0.12634 \dots + πn \leq x \leq 1.69713 \dots + πn

求解步骤

2 sin (2 x) + \frac{1}{2} \geq 0

将

\frac{1}{2}

到右边

2 sin (2 x) + \frac{1}{2} \geq 0

两边减去

\frac{1}{2}

2 sin (2 x) + \frac{1}{2} - \frac{1}{2} \geq 0 - \frac{1}{2}

化简

2 sin (2 x) \geq - \frac{1}{2}

2 sin (2 x) \geq - \frac{1}{2}

两边除以

2

2 sin (2 x) \geq - \frac{1}{2}

两边除以

2

\frac{2 sin ( 2 x )}{2} \geq \frac{- \frac{1}{2}}{2}

化简

\frac{2 sin ( 2 x )}{2} \geq \frac{- \frac{1}{2}}{2}

化简

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

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

数字相除:

\frac{2}{2} = 1

= sin (2 x)

化简

\frac{- \frac{1}{2}}{2} : - \frac{1}{4}

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

使用分式法则:

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

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

使用分式法则:

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

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

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

数字相乘:

2 \cdot 2 = 4

= - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

对于

sin (x) \geq a

，若

- 1 < a < 1

，则

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x and 2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x : x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x

交换两边

2 x \geq arcsin (- \frac{1}{4}) + 2 πn

化简

arcsin (- \frac{1}{4}) + 2 πn : - arcsin (\frac{1}{4}) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{4}) = - arcsin (\frac{1}{4})

= - arcsin (\frac{1}{4}) + 2 πn

2 x \geq - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 x \geq - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} \geq - \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn : x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

化简

π - arcsin (- \frac{1}{4}) + 2 πn : π + arcsin (\frac{1}{4}) + 2 πn

π - arcsin (- \frac{1}{4}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{4}) = - arcsin (\frac{1}{4})

= π - (- arcsin (\frac{1}{4})) + 2 πn

使用法则

- (- a) = a

= π + arcsin (\frac{1}{4}) + 2 πn

2 x \leq π + arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 x \leq π + arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} \leq \frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

x \leq \frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} + πn

化简

\frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} : \frac{π + arcsin ( \frac{1}{4} )}{2}

\frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2}

使用法则

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

= \frac{π + arcsin ( \frac{1}{4} )}{2}

x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

合并区间

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn and x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

合并重叠的区间

- \frac{arcsin ( \frac{1}{4} )}{2} + πn \leq x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

流行的例子

2sin^2(x)-1<= 0

2 sin^{2} (x) - 1 \leq 0

2sin^2(x)+3sin(x)>= 2

2 sin^{2} (x) + 3 sin (x) \geq 2

cos (x) \leq - \frac{1}{2}

cos^2(x)<=-sin(x)

cos^{2} (x) \leq - sin (x)

tan(θ)>0,sin(θ)<0

tan (θ) > 0, sin (θ) < 0