Question

how would I solve: cos^2 x+2cosx-3=0

Answer 1

\[l et~~~~\cos(x)=a\]
\[a^2+2a-3=0\]\[(a-1)(a+3)=0\]\[a-1=0~~~~─── > ~~a=1,~~~or\]\[a+3=0~~~~─── > ~~a=-3.~\]

Answer 2

\[Cos(x)=1~~~~~\cos^{-1}(1)=?\]\[Cos(x)=-3~~~~~\cos^{-1}(-3)=?\]

Answer 3

I don't have a calculator, sorry -:(

Answer 4

you need no calculator for these
there is no way that \(\cos(x)=-3\)

Answer 5

And therefore, @paris32, among your two potential answers you have only one possible answer that could be correct. What is it? a = ?

If a = ?, what is \[\cos ^{-1} (?) ?\] Hint: you can (or should be able to) evaluate\[\cos ^{-1}1\]

Answer 6

Thank you for your help!