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受欢迎的三角函数 >

tan(x)<= cos(x)

解答

tan (x) \leq cos (x)

解答

2 πn \leq x \leq 0.66623 \dots + 2 πn or \frac{π}{2} + 2 πn < x \leq π - 0.66623 \dots + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

+2

间隔符号

[2 πn, 0.66623 \dots + 2 πn] \cup (\frac{π}{2} + 2 πn, π - 0.66623 \dots + 2 πn] \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn]

十进制

2 πn \leq x \leq 0.66623 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.47535 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

求解步骤

tan (x) \leq cos (x)

将

cos (x)

para o lado esquerdo

tan (x) \leq cos (x)

两边减去

cos (x)

tan (x) - cos (x) \leq cos (x) - cos (x)

tan (x) - cos (x) \leq 0

tan (x) - cos (x) \leq 0

tan (x) - cos (x)

的周期:

2 π

周期函数和的复合周期是这些周期的最小公倍数

tan (x), cos (x)

tan (x)

的周期:

π

tan (x)

的周期是

π

= π

cos (x)

的周期:

2 π

cos (x)

的周期是

2 π

= 2 π

合并周期:

π, 2 π

= 2 π

用 sin, cos 表示

tan (x) - cos (x) \leq 0

使用基本三角恒等式:

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

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

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

化简

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

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

将项转换为分式:

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

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

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

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

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

确定

0 \leq x < 2 π

时

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

\frac{sin ( x ) - cos ^{2} ( x )}{cos ( x )} = 0, 0 \leq x < 2 π : x = 0.66623 \dots, x = π - 0.66623 \dots

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

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

用替代法求解

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

令

: sin (x) = u

- 1 + u + u^{2} = 0

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

- 1 + u + u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + u - 1 = 0

使用求根公式求解

u^{2} + u - 1 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

二次方程组的解是:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u = sin (x)

代回

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

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

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

使用反三角函数性质

sin (x) = \frac{- 1 + 5}{2}

sin (x) = \frac{- 1 + 5}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

在

0 \leq x < 2 π

范围内的解

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

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

无解

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

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

以小数形式表示解

x = 0.66623 \dots, x = π - 0.66623 \dots

确定无定义点:

x = \frac{π}{2}, x = \frac{3 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 < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

0.66623 \dots, \frac{π}{2}, π - 0.66623 \dots, \frac{3 π}{2}

确定区间

0 < x < 0.66623 \dots, 0.66623 \dots < x < \frac{π}{2}, \frac{π}{2} < x < π - 0.66623 \dots, π - 0.66623 \dots < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

总结如下表：

sin (x) - cos^{2} (x) cos (x) \frac{s i n ( x ) - c o s ^{2} ( x )}{c o s ( x )} x = 0 - + - 0 < x < 0.66623 \dots - + - x = 0.66623 \dots 0 + 0 0.66623 \dots < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 未定义 \frac{π}{2} < x < π - 0.66623 \dots + - - x = π - 0.66623 \dots 0 - 0 π - 0.66623 \dots < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} - 0 未定义 \frac{3 π}{2} < x < 2 π - + - x = 2 π - + -

确定满足所需条件的区间：

\leq 0

x = 0 or 0 < x < 0.66623 \dots or x = 0.66623 \dots or \frac{π}{2} < x < π - 0.66623 \dots or x = π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π or x = 2 π

合并重叠的区间

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π or x = 2 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

0 < x < 0.66623 \dots

0 \leq x < 0.66623 \dots

两个区间的并集是指存在于任一区间的数的集合

0 \leq x < 0.66623 \dots

or

x = 0.66623 \dots

0 \leq x \leq 0.66623 \dots

两个区间的并集是指存在于任一区间的数的集合

0 \leq x \leq 0.66623 \dots

or

\frac{π}{2} < x < π - 0.66623 \dots

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x < π - 0.66623 \dots

两个区间的并集是指存在于任一区间的数的集合

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x < π - 0.66623 \dots

or

x = π - 0.66623 \dots

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots

两个区间的并集是指存在于任一区间的数的集合

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots

or

\frac{3 π}{2} < x < 2 π

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π

两个区间的并集是指存在于任一区间的数的集合

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π

or

x = 2 π

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x \leq 2 π

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x \leq 2 π

使用周期

tan (x) - cos (x)

2 πn \leq x \leq 0.66623 \dots + 2 πn or \frac{π}{2} + 2 πn < x \leq π - 0.66623 \dots + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

流行的例子

0.86<= cos^{2(5)}((68)/n)

0.86 \leq cos^{2 (5)} (\frac{68}{n})

2sin(x)-1<0,-2pi<= x<= 0

2 sin (x) - 1 < 0, - 2 π \leq x \leq 0

(cos(x))/(1+cos(2x))<0

\frac{cos ( x )}{1 + cos ( 2 x )} < 0

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

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

sin(x)-cos(x)>1

sin (x) - cos (x) > 1