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 >

3cos^2(θ)-cos(θ)=0

解答

3 cos^{2} (θ) - cos (θ) = 0

解答

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

3 cos^{2} (θ) - cos (θ) = 0

用替代法求解

3 cos^{2} (θ) - cos (θ) = 0

令

: cos (θ) = u

3 u^{2} - u = 0

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

3 u^{2} - u = 0

使用求根公式求解

3 u^{2} - u = 0

二次方程求根公式:

若

a = 3, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 3 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 3 \cdot 0 = 0

4 \cdot 3 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 1}{2 \cdot 3}, u_{2} = \frac{- ( - 1 ) - 1}{2 \cdot 3}

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

\frac{- ( - 1 ) + 1}{2 \cdot 3}

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

u = \frac{- ( - 1 ) - 1}{2 \cdot 3} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 3}

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

= \frac{0}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{0}{6}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = \frac{1}{3}, cos (θ) = 0

cos (θ) = \frac{1}{3}, cos (θ) = 0

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

cos (θ) = \frac{1}{3}

使用反三角函数性质

cos (θ) = \frac{1}{3}

cos (θ) = \frac{1}{3}

的通解

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

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

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

cos (θ) = 0

cos (θ) = 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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

合并所有解

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

以小数形式表示解

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

10sin^2(x)+11sin(x)-8=0

10 sin^{2} (x) + 11 sin (x) - 8 = 0

2cos^2(x)+7cos(x)-4=0

2 cos^{2} (x) + 7 cos (x) - 4 = 0

cot(2θ)=(sqrt(3))/3

cot (2 θ) = \frac{3}{3}

4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

-cos(θ)+1=2cos(θ)+3

- cos (θ) + 1 = 2 cos (θ) + 3