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受欢迎的三角函数 >

3/2 cos(x)-2>= cos(x)-7/4

解答

\frac{3}{2} cos (x) - 2 \geq cos (x) - \frac{7}{4}

解答

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

+2

间隔符号

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

十进制

- 1.04719 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn

求解步骤

\frac{3}{2} cos (x) - 2 \geq cos (x) - \frac{7}{4}

令

: u = cos (x)

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

\frac{3}{2} u - 2 \geq u - \frac{7}{4} : u \geq \frac{1}{2}

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

将

2

到右边

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

两边加上

2

\frac{3}{2} u - 2 + 2 \geq u - \frac{7}{4} + 2

化简

\frac{3}{2} u - 2 + 2 \geq u - \frac{7}{4} + 2

化简

\frac{3}{2} u - 2 + 2 : \frac{3}{2} u

\frac{3}{2} u - 2 + 2

同类项相加:

- 2 + 2 \geq 0

= \frac{3}{2} u

化简

u - \frac{7}{4} + 2 : u + \frac{1}{4}

u - \frac{7}{4} + 2

合并分式

2 - \frac{7}{4} : \frac{1}{4}

2 - \frac{7}{4}

将项转换为分式:

2 = \frac{2 \cdot 4}{4}

= \frac{2 \cdot 4}{4} - \frac{7}{4}

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 \cdot 4 - 7}{4}

2 \cdot 4 - 7 = 1

2 \cdot 4 - 7

数字相乘:

2 \cdot 4 = 8

= 8 - 7

数字相减:

8 - 7 = 1

= 1

= \frac{1}{4}

= u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

将

u

para o lado esquerdo

\frac{3}{2} u \geq u + \frac{1}{4}

两边减去

u

\frac{3}{2} u - u \geq u + \frac{1}{4} - u

化简

\frac{3}{2} u - u \geq u + \frac{1}{4} - u

化简

\frac{3}{2} u - u : \frac{1}{2} u

\frac{3}{2} u - u

因式分解出通项

u

= u (\frac{3}{2} - 1)

\frac{3}{2} - 1 = \frac{1}{2}

\frac{3}{2} - 1

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= \frac{3}{2} - \frac{1 \cdot 2}{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{3 - 1 \cdot 2}{2}

3 - 1 \cdot 2 = 1

3 - 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= 3 - 2

数字相减:

3 - 2 = 1

= 1

= \frac{1}{2}

= \frac{1}{2} u

化简

u + \frac{1}{4} - u : \frac{1}{4}

u + \frac{1}{4} - u

同类项相加:

u - u \geq 0

= \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

在两边乘以

2

\frac{1}{2} u \geq \frac{1}{4}

在两边乘以

2

2 \cdot \frac{1}{2} u \geq \frac{1 \cdot 2}{4}

化简

2 \cdot \frac{1}{2} u \geq \frac{1 \cdot 2}{4}

化简

2 \cdot \frac{1}{2} u : u

2 \cdot \frac{1}{2} u

分式相乘:

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

= \frac{1 \cdot 2}{2} u

约分:

2

= u \cdot 1

乘以:

u \cdot 1 = u

= u

化简

\frac{1 \cdot 2}{4} : \frac{1}{2}

\frac{1 \cdot 2}{4}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

u = cos (x)

代回

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

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (\frac{1}{2}) + 2 πn \leq x \leq arccos (\frac{1}{2}) + 2 πn

化简

- arccos (\frac{1}{2}) : - \frac{π}{3}

- arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

化简

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

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

流行的例子

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