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受欢迎的三角函数 >

solvefor t,cos(wt+k)=(20)/((2pif)^2)

解答

求解

t, cos (wt + k) = \frac{20}{( 2 \cdot π \cdot f ) ^{2}}

解答

t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}, t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}

+1

弧度

t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} - \frac{k}{w} + \frac{2 π}{w} n, t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} - \frac{k}{w} + \frac{2 π}{w} n

求解步骤

cos (wt + k) = \frac{20}{( 2 π f ) ^{2}}

化简

\frac{20}{( 2 π f ) ^{2}} : \frac{5}{π ^{2} f ^{2}}

\frac{20}{( 2 π f ) ^{2}}

(2 π f)^{2} = 2^{2} π^{2} f^{2}

(2 π f)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} π^{2} f^{2}

= \frac{20}{2 ^{2} π ^{2} f ^{2}}

分解

20 : 2^{2} \cdot 5

因式分解

20 = 2^{2} \cdot 5

= \frac{2 ^{2} \cdot 5}{2 ^{2} π ^{2} f ^{2}}

约分:

2^{2}

= \frac{5}{π ^{2} f ^{2}}

使用反三角函数性质

cos (wt + k) = \frac{5}{π ^{2} f ^{2}}

cos (wt + k) = \frac{5}{π ^{2} f ^{2}}

的通解

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

wt + k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn, wt + k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

wt + k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn, wt + k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

解

wt + k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn : t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

wt + k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

将

k

到右边

wt + k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

两边减去

k

wt + k - k = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

化简

wt = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

wt = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

两边除以

w; w \neq = 0

wt = arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

两边除以

w; w \neq = 0

\frac{wt}{w} = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

化简

t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

解

wt + k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn : t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

wt + k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

将

k

到右边

wt + k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn

两边减去

k

wt + k - k = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

化简

wt = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

wt = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

两边除以

w; w \neq = 0

wt = - arccos (\frac{5}{π ^{2} f ^{2}}) + 2 πn - k

两边除以

w; w \neq = 0

\frac{wt}{w} = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

化简

t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}; w \neq = 0

t = \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}, t = - \frac{arccos ( \frac{5}{π ^{2} f ^{2}} )}{w} + \frac{2 πn}{w} - \frac{k}{w}

作图

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