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受欢迎的三角函数 >

(1-2sin(x))(2cos(x)+sqrt(3))<= 0

解答

(1 - 2 sin (x)) (2 cos (x) + 3) \leq 0

解答

\frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

+2

间隔符号

[\frac{π}{6} + 2 πn, \frac{7 π}{6} + 2 πn]

十进制

0.52359 \dots + 2 πn \leq x \leq 3.66519 \dots + 2 πn

求解步骤

(1 - 2 sin (x)) (2 cos (x) + 3) \leq 0

(1 - 2 sin (x)) (2 cos (x) + 3)

的周期:

2 π

(1 - 2 sin (x)) (2 cos (x) + 3)

包含以下函数及对应周期:

sin (x)

的周期为

2 π

复合周期为:

= 2 π

要找到零点，将不等式设置为零

(1 - 2 sin (x)) (2 cos (x) + 3) = 0

解

(1 - 2 sin (x)) (2 cos (x) + 3) = 0

求得

0 \leq x < 2 π

(1 - 2 sin (x)) (2 cos (x) + 3) = 0

分别求解每个部分

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

1 - 2 sin (x) = 0, 0 \leq x < 2 π

将

1

到右边

1 - 2 sin (x) = 0

两边减去

1

1 - 2 sin (x) - 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}

2 cos (x) + 3 = 0 : x = \frac{5 π}{6} or x = \frac{7 π}{6}

2 cos (x) + 3 = 0, 0 \leq x < 2 π

将

3

到右边

2 cos (x) + 3 = 0

两边减去

3

2 cos (x) + 3 - 3 = 0 - 3

化简

2 cos (x) = - 3

2 cos (x) = - 3

两边除以

2

2 cos (x) = - 3

两边除以

2

\frac{2 cos ( x )}{2} = \frac{- 3}{2}

化简

cos (x) = - \frac{3}{2}

cos (x) = - \frac{3}{2}

cos (x) = - \frac{3}{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{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{5 π}{6}, x = \frac{7 π}{6}

合并所有解

\frac{π}{6} or \frac{5 π}{6} or \frac{7 π}{6}

零点之间的区间

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < \frac{7 π}{6}, \frac{7 π}{6} < x < 2 π

总结如下表：

1 - 2 sin (x) 2 cos (x) + 3 (1 - 2 sin (x)) (2 cos (x) + 3) x = 0 + + + 0 < x < \frac{π}{6} + + + x = \frac{π}{6} 0 + 0 \frac{π}{6} < x < \frac{5 π}{6} - + - x = \frac{5 π}{6} 000 \frac{5 π}{6} < x < \frac{7 π}{6} + - - x = \frac{7 π}{6} + 00 \frac{7 π}{6} < x < 2 π + + + x = 2 π + + +

确定满足所需条件的区间：

\leq 0

x = \frac{π}{6} or \frac{π}{6} < x < \frac{5 π}{6} or x = \frac{5 π}{6} or \frac{5 π}{6} < x < \frac{7 π}{6} or x = \frac{7 π}{6}

合并重叠的区间

\frac{π}{6} \leq x < \frac{7 π}{6} or x = \frac{7 π}{6}

两个区间的并集是指存在于任一区间的数的集合

x = \frac{π}{6}

or

\frac{π}{6} < x < \frac{5 π}{6}

\frac{π}{6} \leq x < \frac{5 π}{6}

两个区间的并集是指存在于任一区间的数的集合

\frac{π}{6} \leq x < \frac{5 π}{6}

or

x = \frac{5 π}{6}

\frac{π}{6} \leq x \leq \frac{5 π}{6}

两个区间的并集是指存在于任一区间的数的集合

\frac{π}{6} \leq x \leq \frac{5 π}{6}

or

\frac{5 π}{6} < x < \frac{7 π}{6}

\frac{π}{6} \leq x < \frac{7 π}{6}

两个区间的并集是指存在于任一区间的数的集合

\frac{π}{6} \leq x < \frac{7 π}{6}

or

x = \frac{7 π}{6}

\frac{π}{6} \leq x \leq \frac{7 π}{6}

\frac{π}{6} \leq x \leq \frac{7 π}{6}

使用周期

(1 - 2 sin (x)) (2 cos (x) + 3)

\frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

流行的例子

sqrt(3)cos(x)-1<0

3 cos (x) - 1 < 0

tan(x)>-sqrt(3)

tan (x) > - 3

(9sin(t)+1)/8 <= 2pi

\frac{9 sin ( t ) + 1}{8} \leq 2 π

(2cos(θ)+1)/(2sin(θ)-sqrt(3))>0

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

- 1 - cos (t) \geq 0