Solve the equation sin2x+cos2x=1, if 0

We'll recall the double angle identities for sin 2x and
cos 2x:

sin 2x = 2 sinx*cos
x

cos 2x = `cos^(2)` x - `sin^(2)`
x

We'll recall the Pythagorean
identity:

`sin^(2)` x + `cos^(2)` x =
1

We'll re-write the equation in terms of sin x and cos
x:

2sin x*cos x + `cos^(2)` x - `sin^(2)` x = `sin^(2)` x +
`cos^(2)` x

We'll remove like
terms:

2 sin x*cos x - 2`sin^(2)` x =
0

We'll factorize by sin
x:

sin x*(cos x - sin x) =
0

We'll cancel each
factor:

sin x = 0

x = `pi`
(we'll exclude the values 0 and 2`pi` )

We'll cancel the
next factor:

cos x - sin x =
0

-tan x = -1

tan x =
1

The tangent function has positive values within the 1st
and the 3rd quadrants, therefore the values of x are:

x =
`pi` /4

x = `pi` + `pi` /4

x =
5`pi` /4

Therefore, the solutions of the
equation, over the interval (0,2`pi` ), are: {`pi` /4 ; `pi` ; 5`pi`
/4}.