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受欢迎的三角函数 >

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

解答

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

解答

πn \leq x < 0.61547 \dots + πn or - 0.61547 \dots + π + πn < x \leq π + πn

+2

间隔符号

[πn, 0.61547 \dots + πn) \cup (- 0.61547 \dots + π + πn, π + πn]

十进制

πn \leq x < 0.61547 \dots + πn or 2.52611 \dots + πn < x \leq 3.14159 \dots + πn

求解步骤

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

将

sin^{2} (x)

para o lado esquerdo

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

两边减去

sin^{2} (x)

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

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

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

利用以下特性:

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

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

化简

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

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

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

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

2 sin^{2} (x)

的周期:

π

sin^{n} (x)

的周期

= \frac{sin ( x ) 的周期}{2}

，前提是 n 为偶数

sin (x)

的周期:

2 π

sin (x)

的周期是

2 π

= 2 π

\frac{2 π}{2}

化简

π

合并周期:

π, π

= π

因式分解

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

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

将

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

改写为

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

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

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

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

要找到零点，将不等式设置为零

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

解

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

求得

0 \leq x < π

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

分别求解每个部分

cos (x) + 2 sin (x) = 0 : x = - 0.61547 \dots + π

cos (x) + 2 sin (x) = 0, 0 \leq x < π

使用三角恒等式改写

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

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

1 + 2 tan (x) = 0

1 + 2 tan (x) = 0

将

1

到右边

1 + 2 tan (x) = 0

两边减去

1

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

化简

2 tan (x) = - 1

2 tan (x) = - 1

两边除以

2

2 tan (x) = - 1

两边除以

2

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

化简

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

化简

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

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

约分:

2

= tan (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}

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

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

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

使用反三角函数性质

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

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

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{2}{2}) + πn

x = arctan (- \frac{2}{2}) + πn

在

0 \leq x < π

范围内的解

x = - arctan (\frac{2}{2}) + π

以小数形式表示解

x = - 0.61547 \dots + π

cos (x) - 2 sin (x) = 0 : x = 0.61547 \dots

cos (x) - 2 sin (x) = 0, 0 \leq x < π

使用三角恒等式改写

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

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

1 - 2 tan (x) = 0

1 - 2 tan (x) = 0

将

1

到右边

1 - 2 tan (x) = 0

两边减去

1

1 - 2 tan (x) - 1 = 0 - 1

化简

- 2 tan (x) = - 1

- 2 tan (x) = - 1

两边除以

- 2

- 2 tan (x) = - 1

两边除以

- 2

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

化简

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

化简

\frac{- 2 tan ( x )}{- 2} : tan (x)

\frac{- 2 tan ( x )}{- 2}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

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

约分:

2

= tan (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}

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

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

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

使用反三角函数性质

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

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

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{2}{2}) + πn

x = arctan (\frac{2}{2}) + πn

在

0 \leq x < π

范围内的解

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

以小数形式表示解

x = 0.61547 \dots

合并所有解

0.61547 \dots or - 0.61547 \dots + π

零点之间的区间

0 < x < 0.61547 \dots, 0.61547 \dots < x < - 0.61547 \dots + π, - 0.61547 \dots + π < x < π

总结如下表：

cos (x) + 2 sin (x) cos (x) - 2 sin (x) (cos (x) + 2 sin (x)) (cos (x) - 2 sin (x)) x = 0 + + + 0 < x < 0.61547 \dots + + + x = 0.61547 \dots + 00 0.61547 \dots < x < - 0.61547 \dots + π + - - x = - 0.61547 \dots + π 0 - 0 - 0.61547 \dots + π < x < π - - + x = π - - +

确定满足所需条件的区间：

> 0

x = 0 or 0 < x < 0.61547 \dots or - 0.61547 \dots + π < x < π or x = π

合并重叠的区间

0 \leq x < 0.61547 \dots or - 0.61547 \dots + π < 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

0 < x < 0.61547 \dots

0 \leq x < 0.61547 \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 < 0.61547 \dots

or

- 0.61547 \dots + π < x < π

0 \leq x < 0.61547 \dots or - 0.61547 \dots + π < x < 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 < 0.61547 \dots or - 0.61547 \dots + π < x < π

or

x = π

0 \leq x < 0.61547 \dots or - 0.61547 \dots + π < x \leq π

0 \leq x < 0.61547 \dots or - 0.61547 \dots + π < x \leq π

使用周期

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

πn \leq x < 0.61547 \dots + πn or - 0.61547 \dots + π + πn < x \leq π + πn

流行的例子

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

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

cos^2(θ)>= 1/2

cos^{2} (θ) \geq \frac{1}{2}

sin(x)cos(x)<= 1

sin (x) cos (x) \leq 1

cos(θ)>0,cot(θ)<0

cos (θ) > 0, cot (θ) < 0

(cot(x))^2<1,0<= x<= 2pi

(cot (x))^{2} < 1, 0 \leq x \leq 2 π