How do you solve #cos2x + sin^2x = 0# from 0 to 2pi?

1 Answer
Bill K.
Jul 30, 2015

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

Explanation:

One form of the double-angle formula for cosine is #cos(2x)=1-2sin^{2}(x)# (this is not an equation to solve, it's an "identity", meaning it's true for all #x# where it's defined, which is for all #x\in RR#).

Using this identity, we can re-write #cos(2x)+sin^{2}(x)=0# as #1-2sin^{2}(x)+sin^{2}(x)=0#, or #1-sin^{2}(x)=0#, or #sin^{2}(x)=1#.

The only values of #x# where #sin^{2](x)=1# on the interval #[0,2pi]# are #x=pi/2# and #x=(3pi)/2#.

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