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受欢迎的三角函数 >

1+cos(x)+sin(x)>0

解答

1 + cos (x) + sin (x) > 0

解答

- \frac{π}{2} + 2 πn < x < π + 2 πn

+2

间隔符号

(- \frac{π}{2} + 2 πn, π + 2 πn)

十进制

- 1.57079 \dots + 2 πn < x < 3.14159 \dots + 2 πn

求解步骤

1 + cos (x) + sin (x) > 0

利用以下特性:

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

1 + 2 sin (\frac{π}{4} + x) > 0

将

1

到右边

1 + 2 sin (\frac{π}{4} + x) > 0

两边减去

1

1 + 2 sin (\frac{π}{4} + x) - 1 > 0 - 1

化简

2 sin (\frac{π}{4} + x) > - 1

2 sin (\frac{π}{4} + x) > - 1

两边除以

2

2 sin (\frac{π}{4} + x) > - 1

两边除以

2

\frac{2 sin ( \frac{π}{4} + x )}{2} > \frac{- 1}{2}

化简

\frac{2 sin ( \frac{π}{4} + x )}{2} > \frac{- 1}{2}

化简

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

\frac{2 sin ( \frac{π}{4} + x )}{2}

约分:

2

= sin (\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}

sin (\frac{π}{4} + x) > - \frac{2}{2}

sin (\frac{π}{4} + x) > - \frac{2}{2}

sin (\frac{π}{4} + x) > - \frac{2}{2}

对于

sin (x) > a

，若

- 1 \leq a < 1

，则

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

arcsin (- \frac{2}{2}) + 2 πn < (\frac{π}{4} + x) < π - arcsin (- \frac{2}{2}) + 2 πn

若

a < u < b

，则

a < u and u < b

arcsin (- \frac{2}{2}) + 2 πn < \frac{π}{4} + x and \frac{π}{4} + x < π - arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) + 2 πn < \frac{π}{4} + x : x > - \frac{π}{2} + 2 πn

arcsin (- \frac{2}{2}) + 2 πn < \frac{π}{4} + x

交换两边

\frac{π}{4} + x > arcsin (- \frac{2}{2}) + 2 πn

化简

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

arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) = - \frac{π}{4}

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

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2}{2}) = - arcsin (\frac{2}{2})

= - arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{2})

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}

= \frac{π}{4}

= - \frac{π}{4}

= - \frac{π}{4} + 2 πn

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

\frac{π}{4} + x - \frac{π}{4} > - \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} > 0

= x

化简

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

- \frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= - \frac{π}{4} - \frac{π}{4} + 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 πn

x > - \frac{π}{2} + 2 πn

x > - \frac{π}{2} + 2 πn

x > - \frac{π}{2} + 2 πn

\frac{π}{4} + x < π - arcsin (- \frac{2}{2}) + 2 πn : x < π + 2 πn

\frac{π}{4} + x < π - arcsin (- \frac{2}{2}) + 2 πn

化简

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

π - arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) = - \frac{π}{4}

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

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2}{2}) = - arcsin (\frac{2}{2})

= - arcsin (\frac{2}{2})

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{2})

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}

= \frac{π}{4}

= - \frac{π}{4}

= π - (- \frac{π}{4}) + 2 πn

使用法则

- (- a) = a

= π + \frac{π}{4} + 2 πn

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

\frac{π}{4} + x - \frac{π}{4} < 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} < 0

= x

化简

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

π + \frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

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

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= π + 2 πn

x < π + 2 πn

x < π + 2 πn

x < π + 2 πn

合并区间

x > - \frac{π}{2} + 2 πn and x < π + 2 πn

合并重叠的区间

- \frac{π}{2} + 2 πn < x < π + 2 πn

流行的例子

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

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

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