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受欢迎的三角函数 >

3cos^2(θ)+2cos(θ)=0

解答

3 cos^{2} (θ) + 2 cos (θ) = 0

解答

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos^{2} (θ) + 2 cos (θ) = 0

用替代法求解

3 cos^{2} (θ) + 2 cos (θ) = 0

令

: cos (θ) = u

3 u^{2} + 2 u = 0

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

3 u^{2} + 2 u = 0

使用求根公式求解

3 u^{2} + 2 u = 0

二次方程求根公式:

若

a = 3, b = 2, c = 0

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

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

2^{2} - 4 \cdot 3 \cdot 0 = 2

2^{2} - 4 \cdot 3 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

u = \frac{- 2 + 2}{2 \cdot 3} : 0

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

数字相加/相减:

- 2 + 2 = 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

u = \frac{- 2 - 2}{2 \cdot 3} : - \frac{2}{3}

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

数字相减:

- 2 - 2 = - 4

= \frac{- 4}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 4}{6}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{6}

约分:

2

= - \frac{2}{3}

二次方程组的解是:

u = 0, u = - \frac{2}{3}

u = cos (θ)

代回

cos (θ) = 0, cos (θ) = - \frac{2}{3}

cos (θ) = 0, cos (θ) = - \frac{2}{3}

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

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

cos (θ) = - \frac{2}{3}

使用反三角函数性质

cos (θ) = - \frac{2}{3}

cos (θ) = - \frac{2}{3}

的通解

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

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

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

合并所有解

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

以小数形式表示解

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

作图

流行的例子

sin(θ)+1-cos(θ)=0

sin (θ) + 1 - cos (θ) = 0

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

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

2 cos^{2} (2 x) - 1 = 0

5cos(x)cot(x)-3cos(x)-10cot(x)+6=0

5 cos (x) cot (x) - 3 cos (x) - 10 cot (x) + 6 = 0

cos(x)=(-1)/(sqrt(2))

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