Question

how would you solve sin^2x + cosx + 1?

Answer 1

Could we rewrite sin^2x

Answer 2

use the identity\[\sin^2x+\cos^2x=1\]and sub in for \(\sin^2x\)

Answer 3

so (1 - cos^2x) + 1

Answer 4

then combine the 1s?

Answer 5

yeah plus you got that other cosine still there, so you end up with a quadrtic equation (in cosine)

Answer 6

\[\sin^2(x) + \cos(x) + 1\]hold on just noticed there was no equal sign

Answer 7

is it
\[\sin^2(x) + \cos(x) + 1=0\]?

Answer 8

(1 - cos^2x) + cosx + 1, the goal is to simplify I think?

Answer 9

wait, YES!

Answer 10

i don't know why that would be any simpler

Answer 11

it's = 0 @satellite73

Answer 12

whew not it make sense

Answer 13

\[\sin^2(x) + \cos(x) + 1=0\]
\[(1-\cos^2(x)+\cos(x)+1=0\]
\[-\cos^2(x)+\cos(x)+2=0\]
\[\cos^2(x)-\cos(x)-2=0\] solve as you would
\[z^2-z-2=0\] by factoring

Answer 14

thank you so much! we have a test tomorrow, but for some reason by looking at the equations, i've been overcomplicating everything, and forgetting how to solve...

Answer 15

when factoring, don't I have to check for extraneous solutions?

Answer 16

\[(\cos(x)-2)(\cos(x)+1)=0\]
\[\cos(x)=2\] which you can ignore because it never is 2 and
\[\cos(x)=-1\] which you can solve easily

Answer 17

extranous or whatever your job is to solve for \(x\) not \(\cos(x)\) so you still have to complete the problem,
not hard in this case since \(\cos(x)=-1\) at \(x=\pi, 3\pi, 5\pi, ...\)

Answer 18

good luck on your exam!!

Answer 19

alright, thank you so much! you've helped so much!