# `(sin(2x) + cos(2x))^2 = 1` Find the exact solutions of the equation in the interval [0, 2pi).

## Expert Answers

You need to evaluate the solution to the equation `(cos 2x + sin 2x)^2 = 1` , such that:

`cos^2 2x + 2cos 2x*sin 2x + sin^2 2x = 1`

You need to use the formula `cos^2 2x + sin^2 2x = 1` , such that:

`1 + 2cos 2x*sin...

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You need to evaluate the solution to the equation `(cos 2x + sin 2x)^2 = 1` , such that:

`cos^2 2x + 2cos 2x*sin 2x + sin^2 2x = 1`

You need to use the formula `cos^2 2x + sin^2 2x = 1` , such that:

`1 + 2cos 2x*sin 2x = 1`

Reducing like terms yields:

`2cos 2x*sin 2x = 0`

You need to use the double angle formula such that:

`2cos 2x*sin 2x = sin 4x`

`sin 4x = 0 => 4x = 0 => x = 0`

`sin 4x = 0 => 4x = pi => x = pi/4`

Hence, the solution to the equation, in `[0,2pi)` , are `x = 0, x = pi/4.`

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