Solve for x,y?

| sinx +cosx|^(sin^2x-frac{1}4)=1 +|siny|
and cos^2y=1+sin^2y

1 Answer
Mar 16, 2018

.Someone please double check...
see below

Explanation:

Let's try to solve for y in the second equation:
cos^2y=1+sin^2y
(1-sin^2y)=1+sin^2y
sin^2y= 0
y= 0, pi
General solution for y:
y= 0+pin
Where n is an element of all integers

Solve for x:
Plug in any feasible solution for y:
|sinx+cosx|^(sin^2x-1/4)= 1+|sin(0)|
|sinx+cosx|^(sin^2x-1/4)= 1
Recalling:
x^0=1
sin^2x-1/4=0
sin^2x=1/4
sinx= +-1/2
x= pi/6, (5pi)/6, (7pi)/6, (11pi)/6

General solutions for x:
x= pi/6+pin
x= (5pi)/6+pin
Where n is an element of all integers