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solve for x cos^2x-sin^2x=sin x
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\[\cos^2x-\sin^2x =sinx\]
I see trig... I see trig stuff squared... im thinkin trig identities right off the bat!
cox 2x comes to mind...
but it's useless, I think?
no i like the thinking!
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LOL, this is a quadratic equation in disguise
cos^2 x = 1 - sin^2 x right? so we can turn it into... 1 - sin^2 x - sin^2 x = sin x
\[\cos^2x-\sin^2=1-2\sin^2x\]
oops :P ditto
Oh, you can do it from here? great :)
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I'm not the asker @ksaimouli can you take it from here?
derp, I forgot about that (sorry)
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