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受欢迎的三角函数 >

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

解答

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

解答

\frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

+2

间隔符号

[\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn]

十进制

1.57079 \dots + 2 πn \leq x \leq 4.71238 \dots + 2 πn

求解步骤

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

利用以下特性:

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

因此

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

(1 - cos^{2} (x)) cos (x) - cos (x) \geq 0

化简

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = cos (x), b = 1, c = cos^{2} (x)

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

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

化简

1 \cdot cos (x) - cos^{2} (x) cos (x) : cos (x) - cos^{3} (x)

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

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

cos^{2} (x) cos (x) = cos^{3} (x)

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

使用指数法则:

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

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

= cos^{2 + 1} (x)

数字相加:

2 + 1 = 3

= cos^{3} (x)

= cos (x) - cos^{3} (x)

= cos (x) - cos^{3} (x)

= cos (x) - cos^{3} (x) - cos (x)

同类项相加:

cos (x) - cos (x) = 0

= - cos^{3} (x)

- cos^{3} (x) \geq 0

在两边乘以

- 1

- cos^{3} (x) \geq 0

两边乘以 -1（不等式变号）

(- cos^{3} (x)) (- 1) \leq 0 \cdot (- 1)

化简

cos^{3} (x) \leq 0

cos^{3} (x) \leq 0

对于

u^{n} \leq 0

，若

n

为奇数，则

u \leq 0

cos (x) \leq 0

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (0) + 2 πn \leq x \leq 2 π - arccos (0) + 2 πn

化简

arccos (0) : \frac{π}{2}

arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

化简

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{2}

化简

2 π - \frac{π}{2}

将项转换为分式:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

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{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

数字相乘:

2 \cdot 2 = 4

= 4 π - π

同类项相加:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

\frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

流行的例子

(sin(x)+4)(sqrt(3)-2sin(x))>0

(sin (x) + 4) (3 - 2 sin (x)) > 0

6.5<= 5+3sin(30t)

6.5 \leq 5 + 3 sin (30 t)

solvefor x,2*cos(4x)<= 5

so l v e f or x, 2 \cdot cos (4 x) \leq 5

2cos^2(x)+cos(x)-1<= 0

2 cos^{2} (x) + cos (x) - 1 \leq 0

sin(x/3)<(sqrt(3))/2

sin (\frac{x}{3}) < \frac{3}{2}