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Popular Trigonometria >

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

Solução

resolver para

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

Solução

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}

Radianos

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

Passos da solução

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

Simplificar

\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}

Aplicar as propriedades dos expoentes:

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

= 2^{2} π^{2} f^{2}

= \frac{20}{2 ^{2} π ^{2} f ^{2}}

Fatorar

20 : 2^{2} \cdot 5

Fatorar

20 = 2^{2} \cdot 5

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

Eliminar o fator comum:

2^{2}

= \frac{5}{π ^{2} f ^{2}}

Aplique as propriedades trigonométricas inversas

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

Soluções gerais para

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

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

Resolver

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

Mova

k

para o lado direito

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

Subtrair

k

de ambos os lados

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

Simplificar

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

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

Dividir ambos os lados por

w; w \neq = 0

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

Dividir ambos os lados por

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

Simplificar

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

Resolver

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

Mova

k

para o lado direito

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

Subtrair

k

de ambos os lados

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

Simplificar

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

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

Dividir ambos os lados por

w; w \neq = 0

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

Dividir ambos os lados por

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

Simplificar

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}

Gráfico

Exemplos populares

0=-cos(x)(2sin(x)+3)

0 = - cos (x) (2 sin (x) + 3)

28cos(t)=20,0<= t<360

28 cos (t) = 20, 0^{\circ} \leq t < 36 0^{\circ}

8tan(x/2)+8cos(x)tan(x/2)=1

8 tan (\frac{x}{2}) + 8 cos (x) tan (\frac{x}{2}) = 1

sin(a)=0.5937

sin (a) = 0.5937

tan(x)= 18/25

tan (x) = \frac{18}{25}

solvefor t,cos(wt+k)=(20)/((2*pi*f)^2)

Solução

Solução

Gráfico

Exemplos populares

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