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受欢迎的三角函数 >

cos(x)<= 1+sin(x)

解答

cos (x) \leq 1 + sin (x)

解答

2 πn \leq x \leq \frac{3 π}{2} + 2 πn

+2

间隔符号

[2 πn, \frac{3 π}{2} + 2 πn]

十进制

2 πn \leq x \leq 4.71238 \dots + 2 πn

求解步骤

cos (x) \leq 1 + sin (x)

将

sin (x)

para o lado esquerdo

cos (x) \leq 1 + sin (x)

两边减去

sin (x)

cos (x) - sin (x) \leq 1 + sin (x) - sin (x)

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

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

利用以下特性:

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

2 cos (\frac{π}{4} + x) \leq 1

两边除以

2

2 cos (\frac{π}{4} + x) \leq 1

两边除以

2

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

化简

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

化简

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

\frac{2 cos ( \frac{π}{4} + x )}{2}

约分:

2

= cos (\frac{π}{4} + x)

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}

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

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

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

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

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

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

交换两边

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

化简

arccos (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

\frac{π}{4} + x \geq \frac{π}{4} + 2 πn

将

\frac{π}{4}

到右边

\frac{π}{4} + x \geq \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

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

化简

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

化简

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} \geq 0

= x

化简

\frac{π}{4} + 2 πn - \frac{π}{4} : 2 πn

\frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= \frac{π}{4} - \frac{π}{4} + 2 πn

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= 2 πn

x \geq 2 πn

x \geq 2 πn

x \geq 2 πn

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

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

化简

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

2 π - arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{4} + 2 πn

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} \leq 0

= x

化简

2 π - \frac{π}{4} + 2 πn - \frac{π}{4} : - \frac{π}{2} + 2 π + 2 πn

2 π - \frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= - \frac{π}{4} - \frac{π}{4} + 2 π + 2 πn

合并分式

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

使用法则

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

= \frac{- π - π}{4}

同类项相加:

- π - π = - 2 π

= \frac{- 2 π}{4}

使用分式法则:

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

= - \frac{2 π}{4}

约分:

2

= - \frac{π}{2}

= - \frac{π}{2} + 2 π + 2 πn

x \leq - \frac{π}{2} + 2 π + 2 πn

x \leq - \frac{π}{2} + 2 π + 2 πn

x \leq - \frac{π}{2} + 2 π + 2 πn

化简

- \frac{π}{2} + 2 π : \frac{3 π}{2}

- \frac{π}{2} + 2 π

将项转换为分式:

2 π = \frac{2 π 2}{2}

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

- π + 2 π 2 = 3 π

- π + 2 π 2

数字相乘:

2 \cdot 2 = 4

= - π + 4 π

同类项相加:

- π + 4 π = 3 π

= 3 π

= \frac{3 π}{2}

x \leq \frac{3 π}{2} + 2 πn

合并区间

x \geq 2 πn and x \leq \frac{3 π}{2} + 2 πn

合并重叠的区间

2 πn \leq x \leq \frac{3 π}{2} + 2 πn

流行的例子

cos(θ)<0,csc(θ)<0

cos (θ) < 0, csc (θ) < 0

50sin(-(2pi)/3 x-pi/2)>=-15

50 sin (- \frac{2 π}{3} x - \frac{π}{2}) \geq - 15

cos (\frac{π}{6} x) < 0

tan(x)<=-2sqrt(3)

tan (x) \leq - 23

tan(x)<= 1,0<= x<= 2pi

tan (x) \leq 1, 0 \leq x \leq 2 π