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受欢迎的三角函数 >

sin(x)-cos(x)-1>= 0

解答

sin (x) - cos (x) - 1 \geq 0

解答

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

+2

间隔符号

[\frac{π}{2} + 2 πn, π + 2 πn]

十进制

1.57079 \dots + 2 πn \leq x \leq 3.14159 \dots + 2 πn

求解步骤

sin (x) - cos (x) - 1 \geq 0

利用以下特性:

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

- 1 - 2 cos (\frac{π}{4} + x) \geq 0

将

1

到右边

- 1 - 2 cos (\frac{π}{4} + x) \geq 0

两边加上

1

- 1 - 2 cos (\frac{π}{4} + x) + 1 \geq 0 + 1

化简

- 2 cos (\frac{π}{4} + x) \geq 1

- 2 cos (\frac{π}{4} + x) \geq 1

在两边乘以

- 1

- 2 cos (\frac{π}{4} + x) \geq 1

两边乘以 -1（不等式变号）

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

化简

2 cos (\frac{π}{4} + x) \leq - 1

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}

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}

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 + \frac{π}{2}

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{3 π}{4} + 2 πn

arccos (- \frac{2}{2}) + 2 πn

使用以下普通恒等式:

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

\frac{π}{4} + x \geq \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

\frac{π}{4} + x \geq \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

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

化简

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

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

x \geq 2 πn + \frac{π}{2}

x \geq 2 πn + \frac{π}{2}

x \geq 2 πn + \frac{π}{2}

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

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

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

对同类项分组

= 2 π + 2 πn - \frac{π}{4} - \frac{3 π}{4}

合并分式

- \frac{π}{4} - \frac{3 π}{4} : - π

使用法则

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

= \frac{- π - 3 π}{4}

同类项相加:

- π - 3 π = - 4 π

= \frac{- 4 π}{4}

使用分式法则:

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

= - \frac{4 π}{4}

数字相除:

\frac{4}{4} = 1

= - π

= 2 π + 2 πn - π

对同类项分组

= 2 π - π + 2 πn

同类项相加:

2 π - π = π

= π + 2 πn

x \leq π + 2 πn

x \leq π + 2 πn

x \leq π + 2 πn

合并区间

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

合并重叠的区间

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

流行的例子

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

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

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

cos(2x)<1+sin(x)

cos (2 x) < 1 + sin (x)

(1 + sin (2 x)) > 0

cos(x)<= (-1)/2

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