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受欢迎的三角函数 >

(sin(x))/(cos(x))>= 2sin(x)*cos(x)

解答

\frac{sin ( x )}{cos ( x )} \geq 2 sin (x) \cdot cos (x)

解答

\frac{π}{4} + πn \leq x < \frac{π}{2} + πn or \frac{3 π}{4} + πn \leq x \leq π + πn

+2

间隔符号

[\frac{π}{4} + πn, \frac{π}{2} + πn) \cup [\frac{3 π}{4} + πn, π + πn]

十进制

0.78539 \dots + πn \leq x < 1.57079 \dots + πn or 2.35619 \dots + πn \leq x \leq 3.14159 \dots + πn

求解步骤

\frac{sin ( x )}{cos ( x )} \geq 2 sin (x) cos (x)

将

2 sin (x) cos (x)

para o lado esquerdo

\frac{sin ( x )}{cos ( x )} \geq 2 sin (x) cos (x)

两边减去

2 sin (x) cos (x)

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

\frac{sin ( x )}{cos ( x )} - 2 sin (x) cos (x) \geq 0

\frac{sin ( x )}{cos ( x )} - 2 sin (x) cos (x) \geq 0

化简

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

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

将项转换为分式:

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

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

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

sin (x) - 2 sin (x) cos (x) cos (x)

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

2 sin (x) cos (x) cos (x)

使用指数法则:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 sin (x) cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

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

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

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

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

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

的周期:

π

周期函数和的复合周期是这些周期的最小公倍数

\frac{sin ( x )}{cos ( x )}, 2 sin (x) cos (x)

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

的周期:

π

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

包含以下函数及对应周期:

sin (x)

的周期为

2 π

复合周期为:

π

2 sin (x) cos (x)

的周期:

π

2 sin (x) cos (x)

包含以下函数及对应周期:

sin (x)

的周期为

2 π

复合周期为:

π

合并周期:

π, π

= π

确定

0 \leq x < π

时

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

的零点和无定义点

要找到零点，将不等式设置为零

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

\frac{sin ( x ) - 2 cos ^{2} ( x ) sin ( x )}{cos ( x )} = 0, 0 \leq x < π : x = 0, x = \frac{3 π}{4}, x = \frac{π}{4}

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

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

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

分解

sin (x) - 2 cos^{2} (x) sin (x) : - sin (x) (2 cos (x) + 1) (2 cos (x) - 1)

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

因式分解出通项

- sin (x)

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

分解

2 cos^{2} (x) - 1 : (2 cos (x) + 1) (2 cos (x) - 1)

2 cos^{2} (x) - 1

将

2 cos^{2} (x) - 1

改写为

(2 cos (x))^{2} - 1^{2}

2 cos^{2} (x) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos (x))^{2} - 1^{2} = (2 cos (x) + 1) (2 cos (x) - 1)

= (2 cos (x) + 1) (2 cos (x) - 1)

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

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

分别求解每个部分

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

sin (x) = 0, 0 \leq x < π : x = 0

sin (x) = 0, 0 \leq x < 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 < π

范围内的解

x = 0

2 cos (x) + 1 = 0, 0 \leq x < π : x = \frac{3 π}{4}

2 cos (x) + 1 = 0, 0 \leq x < π

将

1

到右边

2 cos (x) + 1 = 0

两边减去

1

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

化简

2 cos (x) = - 1

2 cos (x) = - 1

两边除以

2

2 cos (x) = - 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( x )}{2} : cos (x)

\frac{2 cos ( x )}{2}

约分:

2

= cos (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}

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

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

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

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

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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

在

0 \leq x < π

范围内的解

x = \frac{3 π}{4}

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

2 cos (x) - 1 = 0, 0 \leq x < π

将

1

到右边

2 cos (x) - 1 = 0

两边加上

1

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

化简

2 cos (x) = 1

2 cos (x) = 1

两边除以

2

2 cos (x) = 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( x )}{2} : cos (x)

\frac{2 cos ( x )}{2}

约分:

2

= cos (x)

化简

\frac{1}{2} : \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}

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

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

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

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

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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

在

0 \leq x < π

范围内的解

x = \frac{π}{4}

合并所有解

x = 0, x = \frac{3 π}{4}, x = \frac{π}{4}

确定无定义点:

x = \frac{π}{2}

找到分母的零解

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 < π

范围内的解

x = \frac{π}{2}

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

确定区间

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

总结如下表：

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

确定满足所需条件的区间：

\geq 0

x = 0 or x = \frac{π}{4} or \frac{π}{4} < x < \frac{π}{2} or x = \frac{3 π}{4} or \frac{3 π}{4} < x < π or x = π

合并重叠的区间

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x < π or x = π

两个区间的并集是指存在于任一区间的数的集合

x = 0

or

x = \frac{π}{4}

x = 0 or x = \frac{π}{4}

两个区间的并集是指存在于任一区间的数的集合

x = 0 or x = \frac{π}{4}

or

\frac{π}{4} < x < \frac{π}{2}

x = 0 or \frac{π}{4} \leq x < \frac{π}{2}

两个区间的并集是指存在于任一区间的数的集合

x = 0 or \frac{π}{4} \leq x < \frac{π}{2}

or

x = \frac{3 π}{4}

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or x = \frac{3 π}{4}

两个区间的并集是指存在于任一区间的数的集合

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or x = \frac{3 π}{4}

or

\frac{3 π}{4} < x < π

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x < π

两个区间的并集是指存在于任一区间的数的集合

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x < π

or

x = π

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x \leq π

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x \leq π

使用周期

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

\frac{π}{4} + πn \leq x < \frac{π}{2} + πn or \frac{3 π}{4} + πn \leq x \leq π + πn

流行的例子

tan(x)*tan(2x)>1

tan (x) \cdot tan (2 x) > 1

2cos^3(3x)-cos(3x)<0

2 cos^{3} (3 x) - cos (3 x) < 0

0 \leq sin (π x)

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

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

0.5 \leq sin (30 t)