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受欢迎的三角函数 >

5cos^2(θ)-2cos(θ)=0

解答

5 cos^{2} (θ) - 2 cos (θ) = 0

解答

θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 66.42182 \dots^{\circ} + 36 0^{\circ} n, θ = 293.57817 \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

5 cos^{2} (θ) - 2 cos (θ) = 0

用替代法求解

5 cos^{2} (θ) - 2 cos (θ) = 0

令

: cos (θ) = u

5 u^{2} - 2 u = 0

5 u^{2} - 2 u = 0 : u = \frac{2}{5}, u = 0

5 u^{2} - 2 u = 0

使用求根公式求解

5 u^{2} - 2 u = 0

二次方程求根公式:

若

a = 5, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 2)^{2} - 4 \cdot 5 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 5 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 5 \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 5}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 5}, u_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 5}

u = \frac{- ( - 2 ) + 2}{2 \cdot 5} : \frac{2}{5}

\frac{- ( - 2 ) + 2}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 + 2}{2 \cdot 5}

数字相加:

2 + 2 = 4

= \frac{4}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{4}{10}

约分:

2

= \frac{2}{5}

u = \frac{- ( - 2 ) - 2}{2 \cdot 5} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 - 2}{2 \cdot 5}

数字相减:

2 - 2 = 0

= \frac{0}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{0}{10}

使用法则

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

= 0

二次方程组的解是:

u = \frac{2}{5}, u = 0

u = cos (θ)

代回

cos (θ) = \frac{2}{5}, cos (θ) = 0

cos (θ) = \frac{2}{5}, cos (θ) = 0

cos (θ) = \frac{2}{5} : θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

cos (θ) = \frac{2}{5}

使用反三角函数性质

cos (θ) = \frac{2}{5}

cos (θ) = \frac{2}{5}

的通解

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

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 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{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

以小数形式表示解

θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

tan^2(x)-2tan(x)-3=0

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15cos(x)+8=-cos(x)

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