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受欢迎的三角函数 >

tan(2x)<1

解答

tan (2 x) < 1

解答

- \frac{π}{4} + \frac{π}{2} n < x < \frac{π}{8} + \frac{π}{2} n

+2

间隔符号

(- \frac{π}{4} + \frac{π}{2} n, \frac{π}{8} + \frac{π}{2} n)

十进制

- 0.78539 \dots + \frac{π}{2} n < x < 0.39269 \dots + \frac{π}{2} n

求解步骤

tan (2 x) < 1

若

tan (x) < a

，则

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

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

若

a < u < b

，则

a < u and u < b

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

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

- \frac{π}{2} + πn < 2 x

交换两边

2 x > - \frac{π}{2} + πn

两边除以

2

2 x > - \frac{π}{2} + πn

两边除以

2

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

化简

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

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

x > - \frac{π}{4} + \frac{πn}{2}

x > - \frac{π}{4} + \frac{πn}{2}

x > - \frac{π}{4} + \frac{πn}{2}

2 x < arctan (1) + πn : x < \frac{π}{8} + \frac{πn}{2}

2 x < 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

2 x < \frac{π}{4} + πn

两边除以

2

2 x < \frac{π}{4} + πn

两边除以

2

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

化简

\frac{2 x}{2} < \frac{\frac{π}{4}}{2} + \frac{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}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

x < \frac{π}{8} + \frac{πn}{2}

x < \frac{π}{8} + \frac{πn}{2}

x < \frac{π}{8} + \frac{πn}{2}

合并区间

x > - \frac{π}{4} + \frac{πn}{2} and x < \frac{π}{8} + \frac{πn}{2}

合并重叠的区间

- \frac{π}{4} + \frac{π}{2} n < x < \frac{π}{8} + \frac{π}{2} n

流行的例子

cos(x)<= 1/2 ,-pi<= x<= pi

cos (x) \leq \frac{1}{2}, - π \leq x \leq π

tan (2 x) < 3

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

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

sin (x) < - 0.5

cos^{2} (x) > 14