What is the general solution for solvefor x,cos^2(x)=sin(y+pi/2) ?

The general solution for solvefor x,cos^2(x)=sin(y+pi/2) is

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solvefor x,cos^2(x)=sin(y+pi/2)

solve for

x, cos^{2} (x) = sin (y + \frac{π}{2})

x = arccos (sin (y + \frac{π}{2})) + 2 πn, x = - arccos (sin (y + \frac{π}{2})) + 2 πn, x = arccos (- sin (y + \frac{π}{2})) + 2 πn, x = - arccos (- sin (y + \frac{π}{2})) + 2 πn

Solution steps

cos^{2} (x) = sin (y + \frac{π}{2})

Solve by substitution

cos^{2} (x) = sin (y + \frac{π}{2})

Let:

cos (x) = u

u^{2} = sin (y + \frac{π}{2})

u^{2} = sin (y + \frac{π}{2}) : u = sin (y + \frac{π}{2}), u = - sin (y + \frac{π}{2})

u^{2} = sin (y + \frac{π}{2})

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = sin (y + \frac{π}{2}), u = - sin (y + \frac{π}{2})

Substitute back

u = cos (x)

cos (x) = sin (y + \frac{π}{2}), cos (x) = - sin (y + \frac{π}{2})

cos (x) = sin (y + \frac{π}{2}), cos (x) = - sin (y + \frac{π}{2})

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

cos (x) = sin (y + \frac{π}{2})

Apply trig inverse properties

cos (x) = sin (y + \frac{π}{2})

General solutions for

cos (x) = sin (y + \frac{π}{2})

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

x = arccos (sin (y + \frac{π}{2})) + 2 πn, x = - arccos (sin (y + \frac{π}{2})) + 2 πn

x = arccos (sin (y + \frac{π}{2})) + 2 πn, x = - arccos (sin (y + \frac{π}{2})) + 2 πn

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

cos (x) = - sin (y + \frac{π}{2})

Apply trig inverse properties

cos (x) = - sin (y + \frac{π}{2})

General solutions for

cos (x) = - sin (y + \frac{π}{2})

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

x = arccos (- sin (y + \frac{π}{2})) + 2 πn, x = - arccos (- sin (y + \frac{π}{2})) + 2 πn

x = arccos (- sin (y + \frac{π}{2})) + 2 πn, x = - arccos (- sin (y + \frac{π}{2})) + 2 πn

Combine all the solutions

x = arccos (sin (y + \frac{π}{2})) + 2 πn, x = - arccos (sin (y + \frac{π}{2})) + 2 πn, x = arccos (- sin (y + \frac{π}{2})) + 2 πn, x = - arccos (- sin (y + \frac{π}{2})) + 2 πn

0 = (sin (x))^{2}

sec^{2} (x) - tan^{2} (x) = cos (x)

x, cos^{2} (x) - sin^{2} (x) = 0

cos (x) = \frac{60}{61}, cos (2 x)

x, tan (3 x) = - \frac{1}{3}

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