What is the general solution for solvefor x,13y=cos^4(1-2x) ?

The general solution for solvefor x,13y=cos^4(1-2x) is

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solvefor x,13y=cos^4(1-2x)

solve for

x, 13 y = cos^{4} (1 - 2 x)

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

Solution steps

13 y = cos^{4} (1 - 2 x)

Switch sides

cos^{4} (1 - 2 x) = 13 y

Solve by substitution

cos (1 - 2 x) = 13 y, cos (1 - 2 x) = - 13 y, cos (1 - 2 x) = i 13 y, cos (1 - 2 x) = - i 13 y

cos (1 - 2 x) = 13 y : x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = - 13 y : x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

cos (1 - 2 x) = i 13 y :

No Solution

cos (1 - 2 x) = - i 13 y :

No Solution

Combine all the solutions

x = - \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( 13 y )}{2} - πn + \frac{1}{2}, x = - \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}, x = \frac{arccos ( - 13 y )}{2} - πn + \frac{1}{2}

cos (x) + cos^{2} (x) + cos^{3} (x) = 0

cos (x) - sin (x) = \frac{1}{( sin ( x ) )} - \frac{1}{( cos ( x ) )}

sin^{2} (x) + cos^{5} (x) = 2

16 = 4 + 9 - 12 cos (x)

tanh (z) + 2 = 0

What is the general solution for solvefor x,13y=cos^4(1-2x) ?
The general solution for solvefor x,13y=cos^4(1-2x) is