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受欢迎的三角函数 >

cos(x/2)+sqrt(3)sin(x/2)<0

解答

cos (\frac{x}{2}) + 3 sin (\frac{x}{2}) < 0

解答

\frac{5 π}{3} + 4 πn < x < \frac{11 π}{3} + 4 πn

+2

间隔符号

(\frac{5 π}{3} + 4 πn, \frac{11 π}{3} + 4 πn)

十进制

5.23598 \dots + 4 πn < x < 11.51917 \dots + 4 πn

求解步骤

cos (\frac{x}{2}) + 3 sin (\frac{x}{2}) < 0

令

: u = \frac{x}{2}

cos (u) + 3 sin (u) < 0

cos (u) + 3 sin (u) < 0 : - \frac{7 π}{6} + 2 πn < u < - \frac{π}{6} + 2 πn

cos (u) + 3 sin (u) < 0

使用三角恒等式改写

两边除以

2

\frac{cos ( u ) + 3 sin ( u )}{2} < \frac{0}{2}

\frac{cos ( u ) + 3 sin ( u )}{2} < \frac{0}{2} : \frac{1}{2} cos (u) + \frac{3}{2} sin (u) < 0

\frac{cos ( u ) + 3 sin ( u )}{2} < \frac{0}{2}

乘开

\frac{cos ( u ) + 3 sin ( u )}{2} : \frac{1}{2} cos (u) + \frac{3}{2} sin (u)

\frac{cos ( u ) + 3 sin ( u )}{2}

使用分式法则:

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

\frac{cos ( u ) + 3 sin ( u )}{2} = \frac{cos ( u )}{2} + \frac{3 sin ( u )}{2}

= \frac{cos ( u )}{2} + \frac{3 sin ( u )}{2}

= \frac{1}{2} cos (u) + \frac{3}{2} sin (u)

乘开

\frac{0}{2} : 0

\frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

\frac{1}{2} cos (u) + \frac{3}{2} sin (u) < 0

\frac{1}{2} cos (u) + \frac{3}{2} sin (u) < 0

\frac{3}{2} = cos (\frac{π}{6})

\frac{1}{2} cos (u) + cos (\frac{π}{6}) sin (u) < 0

\frac{1}{2} = sin (\frac{π}{6})

sin (\frac{π}{6}) cos (u) + cos (\frac{π}{6}) sin (u) < 0

利用以下特性:

cos (s) sin (t) + cos (t) sin (s) = sin (s + t)

sin (\frac{π}{6} + u) < 0

sin (\frac{π}{6} + u) < 0

对于

sin (x) < a

，若

- 1 < a \leq 1

，则

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

- π - arcsin (0) + 2 πn < (\frac{π}{6} + u) < arcsin (0) + 2 πn

若

a < u < b

，则

a < u and u < b

- π - arcsin (0) + 2 πn < \frac{π}{6} + u and \frac{π}{6} + u < arcsin (0) + 2 πn

- π - arcsin (0) + 2 πn < \frac{π}{6} + u : u > - \frac{7 π}{6} + 2 πn

- π - arcsin (0) + 2 πn < \frac{π}{6} + u

交换两边

\frac{π}{6} + u > - π - arcsin (0) + 2 πn

化简

- π - arcsin (0) + 2 πn : 2 πn - π

- π - arcsin (0) + 2 πn

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - 0 + 2 πn

- π - 0 + 2 πn = - π + 2 πn

= 2 πn - π

\frac{π}{6} + u > 2 πn - π

将

\frac{π}{6}

到右边

\frac{π}{6} + u > 2 πn - π

两边减去

\frac{π}{6}

\frac{π}{6} + u - \frac{π}{6} > 2 πn - π - \frac{π}{6}

化简

u > 2 πn - π - \frac{π}{6}

u > 2 πn - π - \frac{π}{6}

化简

- π - \frac{π}{6} : - \frac{7 π}{6}

- π - \frac{π}{6}

将项转换为分式:

π = \frac{π 6}{6}

= - \frac{π 6}{6} - \frac{π}{6}

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 6 - π}{6}

同类项相加:

- 6 π - π = - 7 π

= \frac{- 7 π}{6}

使用分式法则:

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

= - \frac{7 π}{6}

u > - \frac{7 π}{6} + 2 πn

\frac{π}{6} + u < arcsin (0) + 2 πn : u < 2 πn - \frac{π}{6}

\frac{π}{6} + u < arcsin (0) + 2 πn

化简

arcsin (0) + 2 πn : 2 πn

arcsin (0) + 2 πn

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

\frac{π}{6} + u < 2 πn

将

\frac{π}{6}

到右边

\frac{π}{6} + u < 2 πn

两边减去

\frac{π}{6}

\frac{π}{6} + u - \frac{π}{6} < 2 πn - \frac{π}{6}

化简

u < 2 πn - \frac{π}{6}

u < 2 πn - \frac{π}{6}

合并区间

u > - \frac{7 π}{6} + 2 πn and u < 2 πn - \frac{π}{6}

合并重叠的区间

- \frac{7 π}{6} + 2 πn < u < - \frac{π}{6} + 2 πn

- \frac{7 π}{6} + 2 πn < u < - \frac{π}{6} + 2 πn

\frac{x}{2} = u

代回

- \frac{7 π}{6} + 2 πn < (\frac{x}{2}) < - \frac{π}{6} + 2 πn

- \frac{7 π}{6} + 2 πn < (\frac{x}{2}) < - \frac{π}{6} + 2 πn : \frac{5 π}{3} + 4 πn < x < \frac{11 π}{3} + 4 πn

- \frac{7 π}{6} + 2 πn < (\frac{x}{2}) < - \frac{π}{6} + 2 πn

若

a < u < b

，则

a < u and u < b

- \frac{7 π}{6} + 2 πn < (\frac{x}{2}) and (\frac{x}{2}) < - \frac{π}{6} + 2 πn

- \frac{7 π}{6} + 2 πn < \frac{x}{2} : x > - \frac{7 π}{3} + 4 πn

- \frac{7 π}{6} + 2 πn < \frac{x}{2}

交换两边

\frac{x}{2} > - \frac{7 π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} > - \frac{7 π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} > - 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} > - 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn : - \frac{7 π}{3} + 4 πn

- 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{6} = \frac{7 π}{3}

2 \cdot \frac{7 π}{6}

分式相乘:

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

= \frac{7 π 2}{6}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{6}

约分:

2

= \frac{7 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= - \frac{7 π}{3} + 4 πn

x > - \frac{7 π}{3} + 4 πn

x > - \frac{7 π}{3} + 4 πn

x > - \frac{7 π}{3} + 4 πn

\frac{x}{2} < - \frac{π}{6} + 2 πn : x < - \frac{π}{3} + 4 πn

\frac{x}{2} < - \frac{π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} < - \frac{π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} < - 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} < - 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 \cdot \frac{π}{6} + 2 \cdot 2 πn : - \frac{π}{3} + 4 πn

- 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 2}{6}

约分:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= - \frac{π}{3} + 4 πn

x < - \frac{π}{3} + 4 πn

x < - \frac{π}{3} + 4 πn

x < - \frac{π}{3} + 4 πn

合并区间

x > - \frac{7 π}{3} + 4 πn and x < - \frac{π}{3} + 4 πn

合并重叠的区间

\frac{5 π}{3} + 4 πn < x < \frac{11 π}{3} + 4 πn

\frac{5 π}{3} + 4 πn < x < \frac{11 π}{3} + 4 πn

流行的例子

solvefor x, 1/(sqrt(3))<= tan(x)

so l v e f or x, \frac{1}{3} \leq tan (x)

sin(θ)cos(θ)tan(θ)-cos^2(θ)>0

sin (θ) cos (θ) tan (θ) - cos^{2} (θ) > 0

- 3 cos (x) + 1 \leq 1

1 - tan (x) \leq 1

sin(x)>sin^2(x)

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