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cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0

解答

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

解答

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

+1

度数

θ = 88.59552 \dots^{\circ} + 36 0^{\circ} n, θ = 271.40447 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

用替代法求解

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 \cdot 1} - 1 = 0

令

: cos (θ) = u

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0 : u = \frac{- 400 + 160384.9444}{19.62}, u = \frac{- 400 - 160384.9444}{19.62}

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0

化简

9.81 \cdot 1 : 9.81

9.81 \cdot 1

数字相乘:

9.81 \cdot 1 = 9.81

= 9.81

u^{2} + \frac{2 0 ^{2} u}{9.81} - 1 = 0

在两边乘以

9.81

u^{2} + \frac{2 0 ^{2} u}{9.81} - 1 = 0

在两边乘以

9.81

u^{2} \cdot 9.81 + \frac{2 0 ^{2} u}{9.81} \cdot 9.81 - 1 \cdot 9.81 = 0 \cdot 9.81

化简

9.81 u^{2} + 400 u - 9.81 = 0

9.81 u^{2} + 400 u - 9.81 = 0

使用求根公式求解

9.81 u^{2} + 400 u - 9.81 = 0

二次方程求根公式:

若

a = 9.81, b = 400, c = - 9.81

u_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 \cdot 9.81 ( - 9.81 )}{2 \cdot 9.81}

u_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 \cdot 9.81 ( - 9.81 )}{2 \cdot 9.81}

40 0^{2} - 4 \cdot 9.81 (- 9.81) = 160384.9444

40 0^{2} - 4 \cdot 9.81 (- 9.81)

使用法则

- (- a) = a

= 40 0^{2} + 4 \cdot 9.81 \cdot 9.81

数字相乘:

4 \cdot 9.81 \cdot 9.81 = 384.9444

= 40 0^{2} + 384.9444

40 0^{2} = 160000

= 160000 + 384.9444

数字相加:

160000 + 384.9444 = 160384.9444

= 160384.9444

u_{1, 2} = \frac{- 400 \pm 160384.9444}{2 \cdot 9.81}

将解分隔开

u_{1} = \frac{- 400 + 160384.9444}{2 \cdot 9.81}, u_{2} = \frac{- 400 - 160384.9444}{2 \cdot 9.81}

u = \frac{- 400 + 160384.9444}{2 \cdot 9.81} : \frac{- 400 + 160384.9444}{19.62}

\frac{- 400 + 160384.9444}{2 \cdot 9.81}

数字相乘:

2 \cdot 9.81 = 19.62

= \frac{- 400 + 160384.9444}{19.62}

u = \frac{- 400 - 160384.9444}{2 \cdot 9.81} : \frac{- 400 - 160384.9444}{19.62}

\frac{- 400 - 160384.9444}{2 \cdot 9.81}

数字相乘:

2 \cdot 9.81 = 19.62

= \frac{- 400 - 160384.9444}{19.62}

二次方程组的解是:

u = \frac{- 400 + 160384.9444}{19.62}, u = \frac{- 400 - 160384.9444}{19.62}

u = cos (θ)

代回

cos (θ) = \frac{- 400 + 160384.9444}{19.62}, cos (θ) = \frac{- 400 - 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62}, cos (θ) = \frac{- 400 - 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62} : θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

使用反三角函数性质

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

的通解

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

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

cos (θ) = \frac{- 400 - 160384.9444}{19.62} :

无解

cos (θ) = \frac{- 400 - 160384.9444}{19.62}

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

以小数形式表示解

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

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