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受欢迎的三角函数 >

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

解答

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

解答

\frac{4 arcsin ( \frac{2}{4} ) - π}{4} + 2 πn \leq x \leq \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

+2

间隔符号

\frac{4 arcsin ( \frac{2}{4} ) - π}{4} + 2 πn, \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

十进制

- 0.42403 \dots + 2 πn \leq x \leq 1.99482 \dots + 2 πn

求解步骤

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

利用以下特性:

cos (x) + sin (x) = 2 sin (\frac{π}{4} + x)

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

两边除以

2

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

两边除以

2

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

化简

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

化简

\frac{2 sin ( \frac{π}{4} + x )}{2} : sin (\frac{π}{4} + x)

\frac{2 sin ( \frac{π}{4} + x )}{2}

约分:

2

= sin (\frac{π}{4} + x)

化简

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

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

使用分式法则:

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

= \frac{1}{2 2}

\frac{1}{2 2}

有理化:

\frac{2}{4}

\frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

= \frac{2}{4}

sin (\frac{π}{4} + x) \geq \frac{2}{4}

sin (\frac{π}{4} + x) \geq \frac{2}{4}

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

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

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

交换两边

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

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

化简

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

arcsin (\frac{2}{4}) - \frac{π}{4}

将项转换为分式:

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

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

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

\frac{π}{4} + x \leq π - arcsin (\frac{2}{4}) + 2 πn : x \leq \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

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

化简

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

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

将项转换为分式:

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

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

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

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

对同类项分组

= 4 π - π - 4 arcsin (\frac{2}{4})

同类项相加:

4 π - π = 3 π

= 3 π - 4 arcsin (\frac{2}{4})

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

x \leq \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

合并区间

x \geq \frac{4 arcsin ( \frac{2}{4} ) - π}{4} + 2 πn and x \leq \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

合并重叠的区间

\frac{4 arcsin ( \frac{2}{4} ) - π}{4} + 2 πn \leq x \leq \frac{3 π - 4 arcsin ( \frac{2}{4} )}{4} + 2 πn

流行的例子

tan(x)<= sqrt(3)

tan (x) \leq 3

50sin(-pi/2 x-pi/2)>= 15

50 sin (- \frac{π}{2} x - \frac{π}{2}) \geq 15

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

so l v e f or x, sin (x) cos (2 x) > 0

cos (2 x) \geq 1

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

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