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受欢迎的三角函数 >

2cos(3x)+cos(2x)+1=0,0<= x<= 2pi

解答

2 cos (3 x) + cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

解答

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0.72273 \dots, x = 2 π - 0.72273 \dots, x = π

+1

度数

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 41.40962 \dots^{\circ}, x = 318.59037 \dots^{\circ}, x = 18 0^{\circ}

求解步骤

2 cos (3 x) + cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

使用三角恒等式改写

1 + cos (2 x) + 2 cos (3 x)

使用倍角公式:

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

= 1 + 2 cos^{2} (x) - 1 + 2 cos (3 x)

化简

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

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

对同类项分组

= 2 cos^{2} (x) + 2 cos (3 x) + 1 - 1

1 - 1 = 0

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

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

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

使用三角恒等式改写

cos (3 x)

改写为

= cos (2 x + x)

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

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

化简

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

乘开

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

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

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

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

2 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)

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

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

1 cos (x)

乘以:

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

= cos (x)

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

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

2 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)

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

化简

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

对同类项分组

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

同类项相加:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

同类项相加:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 2 (4 cos^{3} (x) - 3 cos (x)) + 2 cos^{2} (x)

(- 3 cos (x) + 4 cos^{3} (x)) \cdot 2 + 2 cos^{2} (x) = 0

用替代法求解

(- 3 cos (x) + 4 cos^{3} (x)) \cdot 2 + 2 cos^{2} (x) = 0

令

: cos (x) = u

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0 : u = 0, u = \frac{3}{4}, u = - 1

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0

因式分解

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} : 2 u (4 u - 3) (u + 1)

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2}

因式分解出通项

2

= 2 (u^{3} \cdot 4 - 3 u + u^{2})

分解

4 u^{3} + u^{2} - 3 u : u (4 u - 3) (u + 1)

u^{3} \cdot 4 - 3 u + u^{2}

因式分解出通项

u : u (4 u^{2} + u - 3)

4 u^{3} + u^{2} - 3 u

使用指数法则:

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

u^{2} = uu

= 4 u^{2} u + uu - 3 u

因式分解出通项

u

= u (4 u^{2} + u - 3)

= u (4 u^{2} + u - 3)

分解

4 u^{2} + u - 3 : (4 u - 3) (u + 1)

4 u^{2} + u - 3

改写成标准形式

a x^{2} + b x + c

= 4 u^{2} + u - 3

将表达式拆分成组

4 u^{2} + u - 3

定义

12

的因数:

1, 2, 3, 4, 6, 12

12

约数 (因数)

找到

12

的质因数:

2, 2, 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

乘以

12

的质因数:

4, 6

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6

4, 6

添加质因数:

2, 3

将 1 和数字

12

自身相加

1, 12

12

的因数

1, 2, 3, 4, 6, 12

12

的负因数:

- 1, - 2, - 3, - 4, - 6, - 12

将因数乘以

- 1

得到负因数

- 1, - 2, - 3, - 4, - 6, - 12

对于每两个因数

u * v = - 12

，检验是否

u + v = 1

检验

u = 1, v = - 12 : u * v = - 12, u + v = - 11 \Rightarrow

假检验

u = 2, v = - 6 : u * v = - 12, u + v = - 4 \Rightarrow

假

u = 4, v = - 3

分组为

(a x^{2} + ux) + (vx + c)

(4 u^{2} - 3 u) + (4 u - 3)

= (4 u^{2} - 3 u) + (4 u - 3)

从

4 u^{2} - 3 u

分解出因式

u

:

u (4 u - 3)

4 u^{2} - 3 u

使用指数法则:

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

u^{2} = uu

= 4 uu - 3 u

因式分解出通项

u

= u (4 u - 3)

= u (4 u - 3) + (4 u - 3)

因式分解出通项

4 u - 3

= (4 u - 3) (u + 1)

= u (4 u - 3) (u + 1)

= 2 u (u + 1) (4 u - 3)

= 2 u (4 u - 3) (u + 1)

2 u (4 u - 3) (u + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 4 u - 3 = 0 or u + 1 = 0

解

4 u - 3 = 0 : u = \frac{3}{4}

4 u - 3 = 0

将

3

到右边

4 u - 3 = 0

两边加上

3

4 u - 3 + 3 = 0 + 3

化简

4 u = 3

4 u = 3

两边除以

4

4 u = 3

两边除以

4

\frac{4 u}{4} = \frac{3}{4}

化简

u = \frac{3}{4}

u = \frac{3}{4}

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解为

u = 0, u = \frac{3}{4}, u = - 1

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{3}{4}, cos (x) = - 1

cos (x) = 0, cos (x) = \frac{3}{4}, cos (x) = - 1

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x \leq 2 π

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 \leq 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = \frac{3}{4}, 0 \leq x \leq 2 π : x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4})

cos (x) = \frac{3}{4}, 0 \leq x \leq 2 π

使用反三角函数性质

cos (x) = \frac{3}{4}

cos (x) = \frac{3}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4})

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

在

0 \leq x \leq 2 π

范围内的解

x = π

合并所有解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4}), x = π

以小数形式表示解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0.72273 \dots, x = 2 π - 0.72273 \dots, x = π

作图

流行的例子

tan^2(x)=sqrt(3)

tan^{2} (x) = 3

tan(x)csc^2(x)-2tan(x)=0

tan (x) csc^{2} (x) - 2 tan (x) = 0

5sin(2x)=cos(x)

5 sin (2 x) = cos (x)

2sin(x+pi/3)=-1

2 sin (x + \frac{π}{3}) = - 1

cot^2(a)=cos^2(a)+cos(a)cos(a)

cot^{2} (a) = cos^{2} (a) + cos (a) cos (a)