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受欢迎的三角函数 >

4cos(pi/2+x)+3<= 0

解答

4 cos (\frac{π}{2} + x) + 3 \leq 0

解答

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

+2

间隔符号

[\frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn, \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn]

十进制

0.84806 \dots + 2 πn \leq x \leq 2.29353 \dots + 2 πn

求解步骤

4 cos (\frac{π}{2} + x) + 3 \leq 0

将

3

到右边

4 cos (\frac{π}{2} + x) + 3 \leq 0

两边减去

3

4 cos (\frac{π}{2} + x) + 3 - 3 \leq 0 - 3

化简

4 cos (\frac{π}{2} + x) \leq - 3

4 cos (\frac{π}{2} + x) \leq - 3

两边除以

4

4 cos (\frac{π}{2} + x) \leq - 3

两边除以

4

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

化简

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

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

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

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

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

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

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

交换两边

\frac{π}{2} + x \geq arccos (- \frac{3}{4}) + 2 πn

将

\frac{π}{2}

到右边

\frac{π}{2} + x \geq arccos (- \frac{3}{4}) + 2 πn

两边减去

\frac{π}{2}

\frac{π}{2} + x - \frac{π}{2} \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

化简

x \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

x \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

化简

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

arccos (- \frac{3}{4}) - \frac{π}{2}

将项转换为分式:

arccos (- \frac{3}{4}) = \frac{arccos ( - \frac{3}{4} ) 2}{2}

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

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

x \geq \frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn

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

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

将

\frac{π}{2}

到右边

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

两边减去

\frac{π}{2}

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

化简

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

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

化简

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

2 π - arccos (- \frac{3}{4}) - \frac{π}{2}

将项转换为分式:

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

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

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

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

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

数字相乘:

2 \cdot 2 = 4

= 4 π - 2 arccos (- \frac{3}{4}) - π

对同类项分组

= 4 π - π - 2 arccos (- \frac{3}{4})

同类项相加:

4 π - π = 3 π

= 3 π - 2 arccos (- \frac{3}{4})

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

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

合并区间

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

合并重叠的区间

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

流行的例子

- sin (t) \geq 0

2 cos^{2} (x) < 1

cos (θ) - \frac{1}{2} > 0

solvefor α,cos(α-β)>1

so l v e f or α, cos (α - β) > 1

sin (4 x) > 0.5