zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

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

+2

间隔符号

(- \frac{π}{4} + πn, \frac{π}{4} + πn)

十进制

- 0.78539 \dots + πn < x < 0.78539 \dots + πn

求解步骤

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

将

sin^{2} (x)

para o lado esquerdo

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

两边减去

sin^{2} (x)

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

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

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

利用以下特性:

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

cos (2 x) > 0

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

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

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

若

a < u < b

，则

a < u and u < b

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

- arccos (0) + 2 πn < 2 x : x > - \frac{π}{4} + πn

- arccos (0) + 2 πn < 2 x

交换两边

2 x > - arccos (0) + 2 πn

化简

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

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} + 2 πn

2 x > - \frac{π}{2} + 2 πn

两边除以

2

2 x > - \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= - \frac{π}{4} + πn

x > - \frac{π}{4} + πn

x > - \frac{π}{4} + πn

x > - \frac{π}{4} + πn

2 x < arccos (0) + 2 πn : x < \frac{π}{4} + πn

2 x < arccos (0) + 2 πn

化简

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

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} + 2 πn

2 x < \frac{π}{2} + 2 πn

两边除以

2

2 x < \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x < \frac{π}{4} + πn

x < \frac{π}{4} + πn

x < \frac{π}{4} + πn

合并区间

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

合并重叠的区间

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

流行的例子

sec (2 x) < 0

3/2*cos(2x)>= 0

\frac{3}{2} \cdot cos (2 x) \geq 0

sin^2(x)+1/2 sin(x)>0

sin^{2} (x) + \frac{1}{2} sin (x) > 0

1/2 sin(x)-5<=-1/2

\frac{1}{2} sin (x) - 5 \leq - \frac{1}{2}

2-10cos((pit)/(12))>= 0

2 - 10 cos (\frac{π t}{12}) \geq 0