zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

2 u^{3} - u^{2} = u

2 u^{3} - u^{2} = u : u = 0, u = - \frac{1}{2}, u = 1

2 u^{3} - u^{2} = u

将

u

para o lado esquerdo

2 u^{3} - u^{2} = u

两边减去

u

2 u^{3} - u^{2} - u = u - u

化简

2 u^{3} - u^{2} - u = 0

2 u^{3} - u^{2} - u = 0

因式分解

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

2 u^{3} - u^{2} - u

因式分解出通项

u : u (2 u^{2} - u - 1)

2 u^{3} - u^{2} - u

使用指数法则:

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

u^{2} = uu

= 2 u^{2} u - uu - u

因式分解出通项

u

= u (2 u^{2} - u - 1)

= u (2 u^{2} - u - 1)

分解

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

2 u^{2} - u - 1

将表达式拆分成组

2 u^{2} - u - 1

定义

2

的因数:

1, 2

2

约数 (因数)

找到

2

的质因数:

2

2

2

是质数，因此无法因数分解

= 2

加 1

1

2

的因数

1, 2

2

的负因数:

- 1, - 2

将因数乘以

- 1

得到负因数

- 1, - 2

对于每两个因数

u * v = - 2

，检验是否

u + v = - 1

检验

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

真检验

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

假

u = 1, v = - 2

分组为

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

(2 u^{2} + u) + (- 2 u - 1)

= (2 u^{2} + u) + (- 2 u - 1)

从

2 u^{2} + u

分解出因式

u

:

u (2 u + 1)

2 u^{2} + u

使用指数法则:

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

u^{2} = uu

= 2 uu + u

因式分解出通项

u

= u (2 u + 1)

从

- 2 u - 1

分解出因式

- 1

:

- (2 u + 1)

- 2 u - 1

因式分解出通项

- 1

= - (2 u + 1)

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

因式分解出通项

2 u + 1

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

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 2 u + 1 = 0 or u - 1 = 0

解

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

u = - \frac{1}{2}

u = - \frac{1}{2}

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解为

u = 0, u = - \frac{1}{2}, u = 1

u = cos (x)

代回

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

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

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

cos (x) = 0

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

作图

流行的例子

4cos^2(x)-2=0,0<= x<= 2pi

4 cos^{2} (x) - 2 = 0, 0 \leq x \leq 2 π

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

2 sin^{2} (x) + cos (x) - 1 = 0, 0 \leq x \leq 2 π

sin(x)-tan(x)=cos(x)-1

sin (x) - tan (x) = cos (x) - 1

2 - 8 cos^{2} (x) = 0

sin(2θ-pi/2)=1

sin (2 θ - \frac{π}{2}) = 1