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受欢迎的三角函数 >

(tan(x)-tan^2(x))/(2sin(x)-1)<0

解答

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

解答

2 πn < x < \frac{π}{6} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn or π + 2 πn < x < \frac{5 π}{4} + 2 πn

+2

间隔符号

(2 πn, \frac{π}{6} + 2 πn) \cup (\frac{π}{4} + 2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{5 π}{6} + 2 πn) \cup (π + 2 πn, \frac{5 π}{4} + 2 πn)

十进制

2 πn < x < 0.52359 \dots + 2 πn or 0.78539 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 2.61799 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 3.92699 \dots + 2 πn

求解步骤

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1}

的周期:

2 π

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1}

包含以下函数及对应周期:

tan (x)

的周期为

π

复合周期为:

= 2 π

用 sin, cos 表示

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} < 0

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} < 0

化简

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1} : \frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )}

\frac{\frac{s i n ( x )}{c o s ( x )} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{2 sin ( x ) - 1}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{\frac{s i n ( x )}{c o s ( x )} - \frac{s i n ^{2} ( x )}{c o s ^{2} ( x )}}{2 sin ( x ) - 1}

化简

\frac{sin ( x )}{cos ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

cos (x), cos^{2} (x)

的最小公倍数:

cos^{2} (x)

cos (x), cos^{2} (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

cos^{2} (x)

中的因子组成的表达式

= cos^{2} (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos^{2} (x)

对于

\frac{sin ( x )}{cos ( x )} :

将分母和分子乘以

cos (x)

\frac{sin ( x )}{cos ( x )} = \frac{sin ( x ) cos ( x )}{cos ( x ) cos ( x )} = \frac{sin ( x ) cos ( x )}{cos ^{2} ( x )}

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

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

= \frac{\frac{s i n ( x ) c o s ( x ) - s i n ^{2} ( x )}{c o s ^{2} ( x )}}{2 sin ( x ) - 1}

使用分式法则:

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

= \frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )}

\frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )} < 0

确定

0 \leq x < 2 π

时

\frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )} = 0

\frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )} = 0, 0 \leq x < 2 π : x = 0, x = π, x = \frac{π}{4}, x = \frac{5 π}{4}

\frac{sin ( x ) cos ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) ( 2 sin ( x ) - 1 )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

分解

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

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

使用指数法则:

a^{b + c} = a^{b} a^{c}

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

= sin (x) cos (x) - sin (x) sin (x)

因式分解出通项

sin (x)

= sin (x) (cos (x) - sin (x))

sin (x) (cos (x) - sin (x)) = 0

分别求解每个部分

sin (x) = 0 or cos (x) - sin (x) = 0

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

sin (x) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

在

0 \leq x < 2 π

范围内的解

x = 0, x = π

cos (x) - sin (x) = 0, 0 \leq x < 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

cos (x) - sin (x) = 0, 0 \leq x < 2 π

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{cos ( x ) - sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

1 - \frac{sin ( x )}{cos ( x )} = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{4}, x = \frac{5 π}{4}

合并所有解

x = 0, x = π, x = \frac{π}{4}, x = \frac{5 π}{4}

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

找到分母的零解

cos^{2} (x) (2 sin (x) - 1) = 0

分别求解每个部分

cos^{2} (x) = 0 or 2 sin (x) - 1 = 0

cos^{2} (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos^{2} (x) = 0, 0 \leq x < 2 π

使用法则

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

cos (x) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

2 sin (x) - 1 = 0, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

2 sin (x) - 1 = 0, 0 \leq x < 2 π

将

1

到右边

2 sin (x) - 1 = 0

两边加上

1

2 sin (x) - 1 + 1 = 0 + 1

化简

2 sin (x) = 1

2 sin (x) = 1

两边除以

2

2 sin (x) = 1

两边除以

2

\frac{2 sin ( x )}{2} = \frac{1}{2}

化简

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{6}, x = \frac{5 π}{6}

合并所有解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

0, \frac{π}{6}, \frac{π}{4}, \frac{π}{2}, \frac{5 π}{6}, π, \frac{5 π}{4}, \frac{3 π}{2}

确定区间

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < π, π < x < \frac{5 π}{4}, \frac{5 π}{4} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

总结如下表：

sin (x) cos (x) - sin^{2} (x) cos^{2} (x) 2 sin (x) - 1 \frac{s i n ( x ) c o s ( x ) - s i n ^{2} ( x )}{c o s ^{2} ( x ) ( 2 s i n ( x ) - 1 )} x = 0 0 + - 0 0 < x < \frac{π}{6} + + - - x = \frac{π}{6} + + 0 未定义 \frac{π}{6} < x < \frac{π}{4} + + + + x = \frac{π}{4} 0 + + 0 \frac{π}{4} < x < \frac{π}{2} - + + - x = \frac{π}{2} - 0 + 未定义 \frac{π}{2} < x < \frac{5 π}{6} - + + - x = \frac{5 π}{6} - + 0 未定义 \frac{5 π}{6} < x < π - + - + x = π 0 + - 0 π < x < \frac{5 π}{4} + + - - x = \frac{5 π}{4} 0 + - 0 \frac{5 π}{4} < x < \frac{3 π}{2} - + - + x = \frac{3 π}{2} - 0 - 未定义 \frac{3 π}{2} < x < 2 π - + - + x = 2 π 0 + - 0

确定满足所需条件的区间：

< 0

0 < x < \frac{π}{6} or \frac{π}{4} < x < \frac{π}{2} or \frac{π}{2} < x < \frac{5 π}{6} or π < x < \frac{5 π}{4}

使用周期

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1}

2 πn < x < \frac{π}{6} + 2 πn or \frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn or π + 2 πn < x < \frac{5 π}{4} + 2 πn

流行的例子

sin(x)-cos(x)>1

sin (x) - cos (x) > 1

sin (y) > 0

2 sin (x) + 1 \geq 0

cos^{2} (x) > \frac{5}{6}

((2cos(x)+1))/((2sin(x)-sqrt(3)))>0

\frac{( 2 cos ( x ) + 1 )}{( 2 sin ( x ) - 3 )} > 0