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受欢迎的三角函数 >

sin^4(x)-cos^4(x)<= 0

解答

sin^{4} (x) - cos^{4} (x) \leq 0

解答

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

+2

间隔符号

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

十进制

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

求解步骤

sin^{4} (x) - cos^{4} (x) \leq 0

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

sin^{4} (x), cos^{4} (x)

sin^{4} (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

cos^{4} (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

合并周期:

π, π

= π

因式分解

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

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

将

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

改写为

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

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

使用指数法则:

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

sin^{4} (x) = (sin^{2} (x))^{2}

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

使用指数法则:

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

cos^{4} (x) = (cos^{2} (x))^{2}

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

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

使用平方差公式:

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

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

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

分解

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

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

使用平方差公式:

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

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

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

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

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

要找到零点，将不等式设置为零

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

解

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

求得

0 \leq x < π

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

分别求解每个部分

sin (x) + cos (x) = 0 : x = \frac{3 π}{4}

sin (x) + cos (x) = 0, 0 \leq x < π

使用三角恒等式改写

sin (x) + cos (x) = 0

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

将

1

到右边

tan (x) + 1 = 0

两边减去

1

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

化简

tan (x) = - 1

tan (x) = - 1

tan (x) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

在

0 \leq x < π

范围内的解

x = \frac{3 π}{4}

sin (x) - cos (x) = 0 : x = \frac{π}{4}

sin (x) - cos (x) = 0, 0 \leq x < π

使用三角恒等式改写

sin (x) - cos (x) = 0

在两边除以

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

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

化简

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

使用基本三角恒等式:

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

tan (x) - 1 = 0

tan (x) - 1 = 0

将

1

到右边

tan (x) - 1 = 0

两边加上

1

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

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

在

0 \leq x < π

范围内的解

x = \frac{π}{4}

合并所有解

\frac{π}{4} or \frac{3 π}{4}

零点之间的区间

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

总结如下表：

sin^{2} (x) + cos^{2} (x) sin (x) + cos (x) sin (x) - cos (x) (sin^{2} (x) + cos^{2} (x)) (sin (x) + cos (x)) (sin (x) - cos (x)) x = 0 + + - - 0 < x < \frac{π}{4} + + - - x = \frac{π}{4} + + 00 \frac{π}{4} < x < \frac{3 π}{4} + + + + x = \frac{3 π}{4} + 0 + 0 \frac{3 π}{4} < x < π + - + - x = π + - + -

确定满足所需条件的区间：

\leq 0

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

合并重叠的区间

0 \leq x \leq \frac{π}{4} 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

0 < x < \frac{π}{4}

0 \leq 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

0 \leq x < \frac{π}{4}

or

x = \frac{π}{4}

0 \leq x \leq \frac{π}{4}

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 \frac{π}{4}

or

x = \frac{3 π}{4}

0 \leq x \leq \frac{π}{4} 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

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

or

\frac{3 π}{4} < x < π

0 \leq x \leq \frac{π}{4} 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

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

or

x = π

0 \leq x \leq \frac{π}{4} or \frac{3 π}{4} \leq x \leq π

0 \leq x \leq \frac{π}{4} or \frac{3 π}{4} \leq x \leq π

使用周期

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

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

流行的例子

cos(x)-sin(x)<0

cos (x) - sin (x) < 0

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

sec(-5)sin(θ)>0

sec (- 5) sin (θ) > 0

cos^{2} (x) \geq 1

csc(θ)((sqrt(26))/5)tan(θ)<0

csc (θ) (\frac{26}{5}) tan (θ) < 0