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受欢迎的三角函数 >

15cos^2(θ)+cos(θ)-2=0

解答

15 cos^{2} (θ) + cos (θ) - 2 = 0

解答

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = 1.98231 \dots + 2 πn, θ = - 1.98231 \dots + 2 πn

+1

度数

θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 113.57817 \dots^{\circ} + 36 0^{\circ} n, θ = - 113.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

15 cos^{2} (θ) + cos (θ) - 2 = 0

用替代法求解

15 cos^{2} (θ) + cos (θ) - 2 = 0

令

: cos (θ) = u

15 u^{2} + u - 2 = 0

15 u^{2} + u - 2 = 0 : u = \frac{1}{3}, u = - \frac{2}{5}

15 u^{2} + u - 2 = 0

使用求根公式求解

15 u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 15, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 15 ( - 2 )}{2 \cdot 15}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 15 ( - 2 )}{2 \cdot 15}

1^{2} - 4 \cdot 15 (- 2) = 11

1^{2} - 4 \cdot 15 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 15 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 15 \cdot 2

数字相乘:

4 \cdot 15 \cdot 2 = 120

= 1 + 120

数字相加:

1 + 120 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 1 \pm 11}{2 \cdot 15}

将解分隔开

u_{1} = \frac{- 1 + 11}{2 \cdot 15}, u_{2} = \frac{- 1 - 11}{2 \cdot 15}

u = \frac{- 1 + 11}{2 \cdot 15} : \frac{1}{3}

\frac{- 1 + 11}{2 \cdot 15}

数字相加/相减:

- 1 + 11 = 10

= \frac{10}{2 \cdot 15}

数字相乘:

2 \cdot 15 = 30

= \frac{10}{30}

约分:

10

= \frac{1}{3}

u = \frac{- 1 - 11}{2 \cdot 15} : - \frac{2}{5}

\frac{- 1 - 11}{2 \cdot 15}

数字相减:

- 1 - 11 = - 12

= \frac{- 12}{2 \cdot 15}

数字相乘:

2 \cdot 15 = 30

= \frac{- 12}{30}

使用分式法则:

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

= - \frac{12}{30}

约分:

6

= - \frac{2}{5}

二次方程组的解是:

u = \frac{1}{3}, u = - \frac{2}{5}

u = cos (θ)

代回

cos (θ) = \frac{1}{3}, cos (θ) = - \frac{2}{5}

cos (θ) = \frac{1}{3}, cos (θ) = - \frac{2}{5}

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

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

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

合并所有解

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn, θ = arccos (- \frac{2}{5}) + 2 πn, θ = - arccos (- \frac{2}{5}) + 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 = 1.98231 \dots + 2 πn, θ = - 1.98231 \dots + 2 πn

作图

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