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受欢迎的三角函数 >

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

解答

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

解答

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn or \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

+2

间隔符号

[- \frac{π}{4} + 2 πn, \frac{π}{4} + 2 πn] \cup (\frac{π}{2} + 2 πn, \frac{3 π}{4} + 2 πn] \cup [\frac{5 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn)

十进制

- 0.78539 \dots + 2 πn \leq x \leq 0.78539 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.35619 \dots + 2 πn or 3.92699 \dots + 2 πn \leq x < 4.71238 \dots + 2 πn

求解步骤

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

利用以下特性:

cos^{2} (x) + sin^{2} (x) = 1

因此

sin^{2} (x) = 1 - cos^{2} (x)

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

化简

\frac{2 ( 1 - cos ^{2} ( x ) ) - 1}{cos ( x )} : \frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )}

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

乘开

2 (1 - cos^{2} (x)) - 1 : - 2 cos^{2} (x) + 1

2 (1 - cos^{2} (x)) - 1

乘开

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

2 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

= 2 \cdot 1 - 2 cos^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (x)

= 2 - 2 cos^{2} (x) - 1

化简

2 - 2 cos^{2} (x) - 1 : - 2 cos^{2} (x) + 1

2 - 2 cos^{2} (x) - 1

对同类项分组

= - 2 cos^{2} (x) + 2 - 1

数字相加/相减:

2 - 1 = 1

= - 2 cos^{2} (x) + 1

= - 2 cos^{2} (x) + 1

= \frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )}

\frac{- 2 cos ^{2} ( x ) + 1}{cos ( x )} \leq 0

令

: u = cos (x)

\frac{- 2 u ^{2} + 1}{u} \leq 0

\frac{- 2 u ^{2} + 1}{u} \leq 0 : - \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

\frac{- 2 u ^{2} + 1}{u} \leq 0

分解

\frac{- 2 u ^{2} + 1}{u} : \frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u}

\frac{- 2 u ^{2} + 1}{u}

分解

- 2 u^{2} + 1 : - (2 u + 1) (2 u - 1)

- 2 u^{2} + 1

因式分解出通项

- 1

= - (2 u^{2} - 1)

分解

2 u^{2} - 1 : (2 u + 1) (2 u - 1)

2 u^{2} - 1

将

2 u^{2} - 1

改写为

(2 u)^{2} - 1^{2}

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} u^{2} - 1

将

1

改写为

1^{2}

= (2)^{2} u^{2} - 1^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} u^{2} = (2 u)^{2}

= (2 u)^{2} - 1^{2}

= (2 u)^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 u)^{2} - 1^{2} = (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1)

= - (2 u + 1) (2 u - 1)

= \frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u}

\frac{- ( 2 u + 1 ) ( 2 u - 1 )}{u} \leq 0

两边乘以

- 1

（改变不等式符号）

\frac{( - ( 2 u + 1 ) ( 2 u - 1 ) ) ( - 1 )}{u} \geq 0 \cdot (- 1)

化简

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

确定区间

确定

\frac{( 2 u + 1 ) ( 2 u - 1 )}{u}

符号

确定

2 u + 1

符号

2 u + 1 = 0 : u = - \frac{2}{2}

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

2 u + 1 < 0 : u < - \frac{2}{2}

2 u + 1 < 0

将

1

到右边

2 u + 1 < 0

两边减去

1

2 u + 1 - 1 < 0 - 1

化简

2 u < - 1

2 u < - 1

两边除以

2

2 u < - 1

两边除以

2

\frac{2 u}{2} < \frac{- 1}{2}

化简

\frac{2 u}{2} < \frac{- 1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u < - \frac{2}{2}

u < - \frac{2}{2}

u < - \frac{2}{2}

2 u + 1 > 0 : u > - \frac{2}{2}

2 u + 1 > 0

将

1

到右边

2 u + 1 > 0

两边减去

1

2 u + 1 - 1 > 0 - 1

化简

2 u > - 1

2 u > - 1

两边除以

2

2 u > - 1

两边除以

2

\frac{2 u}{2} > \frac{- 1}{2}

化简

\frac{2 u}{2} > \frac{- 1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u > - \frac{2}{2}

u > - \frac{2}{2}

u > - \frac{2}{2}

确定

2 u - 1

符号

2 u - 1 = 0 : u = \frac{2}{2}

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

\frac{2 u}{2} = \frac{1}{2}

化简

\frac{2 u}{2} = \frac{1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

2 u - 1 < 0 : u < \frac{2}{2}

2 u - 1 < 0

将

1

到右边

2 u - 1 < 0

两边加上

1

2 u - 1 + 1 < 0 + 1

化简

2 u < 1

2 u < 1

两边除以

2

2 u < 1

两边除以

2

\frac{2 u}{2} < \frac{1}{2}

化简

\frac{2 u}{2} < \frac{1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u < \frac{2}{2}

u < \frac{2}{2}

u < \frac{2}{2}

2 u - 1 > 0 : u > \frac{2}{2}

2 u - 1 > 0

将

1

到右边

2 u - 1 > 0

两边加上

1

2 u - 1 + 1 > 0 + 1

化简

2 u > 1

2 u > 1

两边除以

2

2 u > 1

两边除以

2

\frac{2 u}{2} > \frac{1}{2}

化简

\frac{2 u}{2} > \frac{1}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

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}

u > \frac{2}{2}

u > \frac{2}{2}

u > \frac{2}{2}

确定

u

符号

u = 0

u < 0

u > 0

找到奇点

找到分母的零解

u : u = 0

总结如下表：

2 u + 1 2 u - 1 u \frac{( 2 u + 1 ) ( 2 u - 1 )}{u} u < - \frac{2}{2} - - - - u = - \frac{2}{2} 0 - - 0 - \frac{2}{2} < u < 0 + - - + u = 0 + - 0 未定义 0 < u < \frac{2}{2} + - + - u = \frac{2}{2} + 0 + 0 u > \frac{2}{2} + + + +

确定满足所需条件的区间：

\geq 0

u = - \frac{2}{2} or - \frac{2}{2} < u < 0 or u = \frac{2}{2} or u > \frac{2}{2}

合并重叠的区间

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2} or u > \frac{2}{2}

两个区间的并集是指存在于任一区间的数的集合

u = - \frac{2}{2}

or

- \frac{2}{2} < u < 0

- \frac{2}{2} \leq u < 0

两个区间的并集是指存在于任一区间的数的集合

- \frac{2}{2} \leq u < 0

or

u = \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2}

两个区间的并集是指存在于任一区间的数的集合

- \frac{2}{2} \leq u < 0 or u = \frac{2}{2}

or

u > \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

- \frac{2}{2} \leq u < 0 or u \geq \frac{2}{2}

u = cos (x)

代回

- \frac{2}{2} \leq cos (x) < 0 or cos (x) \geq \frac{2}{2}

- \frac{2}{2} \leq cos (x) < 0 : \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

- \frac{2}{2} \leq cos (x) < 0

若

a \leq u < b

，则

a \leq u and u < b

- \frac{2}{2} \leq cos (x) and cos (x) < 0

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

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

交换两边

cos (x) \geq - \frac{2}{2}

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

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

化简

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

- arccos (- \frac{2}{2})

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}

化简

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

arccos (- \frac{2}{2})

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}

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

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (0) + 2 πn < x < 2 π - arccos (0) + 2 πn

化简

arccos (0) : \frac{π}{2}

arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

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}{2}

化简

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

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}{2}

化简

2 π - \frac{π}{2}

将项转换为分式:

2 π = \frac{2 π 2}{2}

= \frac{2 π 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 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

数字相乘:

2 \cdot 2 = 4

= 4 π - π

同类项相加:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

\frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

合并区间

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

合并重叠的区间

\frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

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

cos (x) \geq \frac{2}{2}

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

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

化简

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

- arccos (\frac{2}{2})

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}

化简

arccos (\frac{2}{2}) : \frac{π}{4}

arccos (\frac{2}{2})

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}

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

合并区间

(\frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn) or - \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

合并重叠的区间

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn or \frac{π}{2} + 2 πn < x \leq \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn \leq x < \frac{3 π}{2} + 2 πn

流行的例子

8 sin^{3} (t) < 0

pi/2-arctan(e^x)>0.1

\frac{π}{2} - arctan (e^{x}) > 0.1

arctan (\frac{x}{y}) > 0

cos^2(x)-sin^2(x)>0

cos^{2} (x) - sin^{2} (x) > 0

cos^2(x)>sin^2(x)

cos^{2} (x) > sin^{2} (x)