Find all solutions of the equation in the interval [0,2pi) algebraically. Use the table feature of a graphing utility to check your answers numerically.

sin^2x + cos x + 1=0

Question

Find all solutions of the equation in the interval [0,2pi) algebraically. Use the table feature of a graphing utility to check your answers numerically. 

sin^2x + cos x + 1=0

anonymous · Answer

Does the identity sin^2(x) =1-cos^2(x) help you with anything

anonymous · Answer

no haha

jdoe0001 · Answer

$\bf sin^2(x) + cos(x) + 1=0\
\textrm{as Ahmad1 said, using }sin^2(x) = 1-cos^2(x)\
1-cos^2(x)+ cos(x) + 1=0 \implies -cos^2(x)+ cos(x) +2 = 0$

jdoe0001 · Answer

so you have a quadratic equation

anonymous · Answer

and now you can substitute cos(x)=u and solve for u , and substitute back after that to, get the answer in terms of cosine , but just make sure that x stays in the interval you stated..

anonymous · Answer

ummmmm can someone just write the answer lol

anonymous · Answer

(cos(x)-2)(cos(x)+1)=0
either cos(x)=2 and that's not possible,
or cos(x)=-1 then x=pi

anonymous · Answer

I just factored the expression ..