How do you solve $Cos 2x + sin^2 x = 0?$ from 0 to 2pi?

Question

8623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-03T18:17:52

$x=pi/2,(3pi)/2$

Another method is to use the identity

$cos2x=1-2sin^2x$

So,

$cos2x+sin^2x=0" "=>" "1-2sin^2x+sin^2x=0$

This simplifies to be

$sin^2x=1$

Which implies that, after taking the square root of both sides:

$sinx=+-1$

The only times this occurs on $[0,2pi]$ are at $x=pi/2$ and $x=(3pi)/2$ .

How do you solve Cos 2x + sin^2 x = 0?Cos 2x + sin^2 x = 0? from 0 to 2pi?