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受欢迎的三角函数 >

tan(5x)<= 1

解答

tan (5 x) \leq 1

解答

- \frac{π}{10} + \frac{π}{5} n < x \leq \frac{π}{20} + \frac{π}{5} n

+2

间隔符号

(- \frac{π}{10} + \frac{π}{5} n, \frac{π}{20} + \frac{π}{5} n]

十进制

- 0.31415 \dots + \frac{π}{5} n < x \leq 0.15707 \dots + \frac{π}{5} n

求解步骤

tan (5 x) \leq 1

若

tan (x) \leq a

，则

- \frac{π}{2} + πn < x \leq arctan (a) + πn

- \frac{π}{2} + πn < 5 x \leq arctan (1) + πn

若

a < u \leq b

，则

a < u and u \leq b

- \frac{π}{2} + πn < 5 x and 5 x \leq arctan (1) + πn

- \frac{π}{2} + πn < 5 x : x > - \frac{π}{10} + \frac{πn}{5}

- \frac{π}{2} + πn < 5 x

交换两边

5 x > - \frac{π}{2} + πn

两边除以

5

5 x > - \frac{π}{2} + πn

两边除以

5

\frac{5 x}{5} > - \frac{\frac{π}{2}}{5} + \frac{πn}{5}

化简

\frac{5 x}{5} > - \frac{\frac{π}{2}}{5} + \frac{πn}{5}

化简

\frac{5 x}{5} : x

\frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

- \frac{\frac{π}{2}}{5} + \frac{πn}{5} : - \frac{π}{10} + \frac{πn}{5}

- \frac{\frac{π}{2}}{5} + \frac{πn}{5}

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

使用分式法则:

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

= \frac{π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{π}{10}

= - \frac{π}{10} + \frac{πn}{5}

x > - \frac{π}{10} + \frac{πn}{5}

x > - \frac{π}{10} + \frac{πn}{5}

x > - \frac{π}{10} + \frac{πn}{5}

5 x \leq arctan (1) + πn : x \leq \frac{π}{20} + \frac{πn}{5}

5 x \leq arctan (1) + πn

化简

arctan (1) + πn : \frac{π}{4} + πn

arctan (1) + πn

使用以下普通恒等式:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4} + πn

5 x \leq \frac{π}{4} + πn

两边除以

5

5 x \leq \frac{π}{4} + πn

两边除以

5

\frac{5 x}{5} \leq \frac{\frac{π}{4}}{5} + \frac{πn}{5}

化简

\frac{5 x}{5} \leq \frac{\frac{π}{4}}{5} + \frac{πn}{5}

化简

\frac{5 x}{5} : x

\frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

\frac{\frac{π}{4}}{5} + \frac{πn}{5} : \frac{π}{20} + \frac{πn}{5}

\frac{\frac{π}{4}}{5} + \frac{πn}{5}

\frac{\frac{π}{4}}{5} = \frac{π}{20}

\frac{\frac{π}{4}}{5}

使用分式法则:

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

= \frac{π}{4 \cdot 5}

数字相乘:

4 \cdot 5 = 20

= \frac{π}{20}

= \frac{π}{20} + \frac{πn}{5}

x \leq \frac{π}{20} + \frac{πn}{5}

x \leq \frac{π}{20} + \frac{πn}{5}

x \leq \frac{π}{20} + \frac{πn}{5}

合并区间

x > - \frac{π}{10} + \frac{πn}{5} and x \leq \frac{π}{20} + \frac{πn}{5}

合并重叠的区间

- \frac{π}{10} + \frac{π}{5} n < x \leq \frac{π}{20} + \frac{π}{5} n

流行的例子

(-1)/(16)sec^3(t)>0

\frac{- 1}{16} sec^{3} (t) > 0

cos(x)<= sin(x)

cos (x) \leq sin (x)

2 sin (t) > 0

1 - tan (x) \leq 0

sin(x+23)cos(x-37)>(sqrt(3))/2

sin (x + 2 3^{\circ}) cos (x - 3 7^{\circ}) > \frac{3}{2}