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

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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#.