What is the general solution for solvefor x,y=3cos(fxx+pi/2)+5 ?

The general solution for solvefor x,y=3cos(fxx+pi/2)+5 is

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solvefor x,y=3cos(fxx+pi/2)+5

solve for

x, y = 3 cos (f xx + \frac{π}{2}) + 5

x = \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}

Solution steps

y = 3 cos (f xx + \frac{π}{2}) + 5

Switch sides

3 cos (f xx + \frac{π}{2}) + 5 = y

Move

5

to the right side

3 cos (f xx + \frac{π}{2}) = y - 5

Divide both sides by

3

cos (f xx + \frac{π}{2}) = \frac{y - 5}{3}

Apply trig inverse properties

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

Solve

f xx + \frac{π}{2} = arccos (\frac{y - 5}{3}) + 2 πn : x = \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}; f \neq = 0

Solve

f xx + \frac{π}{2} = - arccos (\frac{y - 5}{3}) + 2 πn : x = \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}; f \neq = 0

x = \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}, x = - \frac{- 2 arccos ( \frac{y - 5}{3} ) + 4 πn - π}{2 f}

sin (x) cos (x) = sin (x), 0 < x \leq 2 π

cos^{5} (x) - sin (x) = 0

x, y = \frac{1}{5 sin ( 2 x )}

cos (θ) = \frac{3}{35}

csc^{2} (θ) - 3 csc (θ) + 2 = 0

What is the general solution for solvefor x,y=3cos(fxx+pi/2)+5 ?
The general solution for solvefor x,y=3cos(fxx+pi/2)+5 is