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受欢迎的三角函数 >

4sin^2(x)+3tan(x)>sec^2(x)

解答

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

解答

\frac{π}{12} + πn < x < \frac{5 π}{12} + πn

+2

间隔符号

(\frac{π}{12} + πn, \frac{5 π}{12} + πn)

十进制

0.26179 \dots + πn < x < 1.30899 \dots + πn

求解步骤

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

将

sec^{2} (x)

para o lado esquerdo

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

两边减去

sec^{2} (x)

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > sec^{2} (x) - sec^{2} (x)

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > 0

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > 0

利用以下特性:

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

因此

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

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x) > 0

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (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

4 (1 - cos^{2} (x)), 3 tan (x), sec^{2} (x)

4 (1 - cos^{2} (x))

的周期:

π

cos^{n} (x)

的周期

= \frac{cos ( x ) 的周期}{2}

，前提是 n 为偶数

cos (x)

的周期:

2 π

cos (x)

的周期是

2 π

= 2 π

\frac{2 π}{2}

化简

π

3 tan (x)

的周期:

π

周期

t an (b x + c) + d = \frac{tan ( x ) 的周期}{∣ b ∣}

tan (x)

的周期是

π

= \frac{π}{∣ 1 ∣}

化简

= π

sec^{2} (x)

的周期:

π

sec^{n} (x)

的周期

= \frac{sec ( x ) 的周期}{2}

，前提是 n 为偶数

sec (x)

的周期:

2 π

sec (x)

的周期是

2 π

= 2 π

\frac{2 π}{2}

化简

π

合并周期:

π, π, π

= π

用 sin, cos 表示

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x) > 0

使用基本三角恒等式:

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

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - sec^{2} (x) > 0

使用基本三角恒等式:

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

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} > 0

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} > 0

化简

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} : \frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )}

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2}

3 \cdot \frac{sin ( x )}{cos ( x )} = \frac{3 sin ( x )}{cos ( x )}

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

分式相乘:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= 4 (- cos^{2} (x) + 1) + \frac{3 sin ( x )}{cos ( x )} - \frac{1}{cos ^{2} ( x )}

将项转换为分式:

4 (- cos^{2} (x) + 1) = \frac{4 ( - cos ^{2} ( x ) + 1 )}{1}

= \frac{4 ( 1 - cos ^{2} ( x ) )}{1} + \frac{sin ( x ) \cdot 3}{cos ( x )} - \frac{1}{cos ^{2} ( x )}

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

的最小公倍数:

cos^{2} (x)

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

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现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{4 ( 1 - cos ^{2} ( x ) )}{1} :

将分母和分子乘以

cos^{2} (x)

\frac{4 ( 1 - cos ^{2} ( x ) )}{1} = \frac{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x )}{1 \cdot cos ^{2} ( x )} = \frac{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x )}{cos ^{2} ( x )}

对于

\frac{sin ( x ) \cdot 3}{cos ( x )} :

将分母和分子乘以

cos (x)

\frac{sin ( x ) \cdot 3}{cos ( x )} = \frac{sin ( x ) \cdot 3 cos ( x )}{cos ( x ) cos ( x )} = \frac{sin ( x ) \cdot 3 cos ( x )}{cos ^{2} ( x )}

= \frac{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ( x ) \cdot 3 cos ( x )}{cos ^{2} ( x )} - \frac{1}{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{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x ) + sin ( x ) \cdot 3 cos ( x ) - 1}{cos ^{2} ( x )}

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} > 0

确定

0 \leq x < π

时

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} = 0

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} = 0, 0 \leq x < π : x = \frac{π}{12}, x = \frac{5 π}{12}

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

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

4 cos^{2} (x) (1 - cos^{2} (x)) + 3 sin (x) cos (x) - 1 = 0

使用三角恒等式改写

- 1 + (1 - cos^{2} (x)) \cdot 4 cos^{2} (x) + 3 cos (x) sin (x)

使用毕达哥拉斯恒等式:

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

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

= - 1 + 3 cos (x) sin (x) + 4 cos^{2} (x) sin^{2} (x)

- 1 + 3 cos (x) sin (x) + 4 cos^{2} (x) sin^{2} (x) = 0

分解

- 1 + 3 cos (x) sin (x) + 4 cos^{2} (x) sin^{2} (x) : (4 sin (x) cos (x) - 1) (sin (x) cos (x) + 1)

- 1 + 3 cos (x) sin (x) + 4 cos^{2} (x) sin^{2} (x)

将表达式拆分成组

4 sin^{2} (x) cos^{2} (x) + 3 sin (x) cos (x) - 1

定义

4

的因数:

1, 2, 4

4

约数 (因数)

找到

4

的质因数:

2, 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2

添加质因数:

2

将 1 和数字

4

自身相加

1, 4

4

的因数

1, 2, 4

4

的负因数:

- 1, - 2, - 4

将因数乘以

- 1

得到负因数

- 1, - 2, - 4

对于每两个因数

u * v = - 4

，检验是否

u + v = 3

检验

u = 1, v = - 4 : u * v = - 4, u + v = - 3 \Rightarrow

假检验

u = 2, v = - 2 : u * v = - 4, u + v = 0 \Rightarrow

假

u = 4, v = - 1

分组为

(a x^{2} y^{2} + ux y) + (vx y + c)

(4 sin^{2} (x) cos^{2} (x) - sin (x) cos (x)) + (4 sin (x) cos (x) - 1)

= (4 sin^{2} (x) cos^{2} (x) - sin (x) cos (x)) + (4 sin (x) cos (x) - 1)

从

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

分解出因式

sin (x) cos (x)

:

sin (x) cos (x) (4 sin (x) cos (x) - 1)

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

使用指数法则:

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

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

= 4 sin (x) sin (x) cos (x) cos (x) - sin (x) cos (x)

因式分解出通项

sin (x) cos (x)

= sin (x) cos (x) (4 sin (x) cos (x) - 1)

= sin (x) cos (x) (4 sin (x) cos (x) - 1) + (4 sin (x) cos (x) - 1)

因式分解出通项

4 sin (x) cos (x) - 1

= (4 sin (x) cos (x) - 1) (sin (x) cos (x) + 1)

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

分别求解每个部分

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

4 sin (x) cos (x) - 1 = 0, 0 \leq x < π : x = \frac{π}{12}, x = \frac{5 π}{12}

4 sin (x) cos (x) - 1 = 0, 0 \leq x < π

使用三角恒等式改写

4 sin (x) cos (x) - 1

使用倍角公式:

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

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

= - 1 + 4 \cdot \frac{sin ( 2 x )}{2}

- 1 + 4 \cdot \frac{sin ( 2 x )}{2} = 0

4 \cdot \frac{sin ( 2 x )}{2} = 2 sin (2 x)

4 \cdot \frac{sin ( 2 x )}{2}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 sin (2 x)

- 1 + 2 sin (2 x) = 0

将

1

到右边

- 1 + 2 sin (2 x) = 0

两边加上

1

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

化简

2 sin (2 x) = 1

2 sin (2 x) = 1

两边除以

2

2 sin (2 x) = 1

两边除以

2

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

化简

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

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

sin (2 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}

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

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

解

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 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}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 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{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

在

0 \leq x < π

范围内的解

x = \frac{π}{12}, x = \frac{5 π}{12}

sin (x) cos (x) + 1 = 0, 0 \leq x < π :

无解

sin (x) cos (x) + 1 = 0, 0 \leq x < π

使用三角恒等式改写

sin (x) cos (x) + 1

使用倍角公式:

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

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

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

1 + \frac{sin ( 2 x )}{2} = 0

将

1

到右边

1 + \frac{sin ( 2 x )}{2} = 0

两边减去

1

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

化简

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

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

在两边乘以

2

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

在两边乘以

2

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

化简

sin (2 x) = - 2

sin (2 x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{π}{12}, x = \frac{5 π}{12}

确定无定义点:

x = \frac{π}{2}

找到分母的零解

cos^{2} (x) = 0

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 < π

范围内的解

x = \frac{π}{2}

\frac{π}{12}, \frac{5 π}{12}, \frac{π}{2}

确定区间

0 < x < \frac{π}{12}, \frac{π}{12} < x < \frac{5 π}{12}, \frac{5 π}{12} < x < \frac{π}{2}, \frac{π}{2} < x < π

总结如下表：

4 cos^{2} (x) (1 - cos^{2} (x)) + 3 sin (x) cos (x) - 1 cos^{2} (x) \frac{4 c o s ^{2} ( x ) ( 1 - c o s ^{2} ( x ) ) + 3 s i n ( x ) c o s ( x ) - 1}{c o s ^{2} ( x )} x = 0 - + - 0 < x < \frac{π}{12} - + - x = \frac{π}{12} 0 + 0 \frac{π}{12} < x < \frac{5 π}{12} + + + x = \frac{5 π}{12} 0 + 0 \frac{5 π}{12} < x < \frac{π}{2} - + - x = \frac{π}{2} - 0 未定义 \frac{π}{2} < x < π - + - x = π - + -

确定满足所需条件的区间：

> 0

\frac{π}{12} < x < \frac{5 π}{12}

使用周期

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x)

\frac{π}{12} + πn < x < \frac{5 π}{12} + πn

流行的例子

sin (4 x + 1 7^{\circ}) > 0

3sin((pix)/(12)-pi/2)<=-2

3 sin (\frac{π x}{12} - \frac{π}{2}) \leq - 2

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

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

1/(sqrt(3))<tan(x)

\frac{1}{3} < tan (x)

6 cos (θ) \geq 0