# Solve the equation sin2x+cos2x=1, if 0<x<2pi?

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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}.**