Question

How do you solve sin^2x+cosx=2

Answer 1

So far I have replaced sin^2x with 1-cos^2x

Answer 2

rewrite \(\sin^2(x)\) as \(1-\cos^2(x)\) so you will have nothing but cosines
then solve the quadratic equation in cosine
then solve for \(x\)

Answer 3

Right now I have cos^2x+cosx+1=0
and I tried to complete the square but it didn't work out

Answer 4

-cosx actually

Answer 5

\[1-\cos^2(x)+\cos(x)=2\]
\[\cos^2(x)-\cos(x)+1=0\]

Answer 6

Yeah that's what I have

Answer 7

no real solutions for sure

Answer 8

i.e. \(x^2-x+1=0\) has no real solutions, so either there is a typo in the problem or they weren't paying attention when they wrote it

Answer 9

Ok thanks because I got a negative under a radical :)