Question

sin2x+cosx=0

Answer 1

start with the fact that \(\sin^2(x)=1-\cos^2(x)\) to rewrite this as
\[1-\cos^2(x)+\cos(x)=0\] or \[\cos^2(x)-\cos(x)+1=0\] then solve the quadratic equation in cosine

Answer 2

ok that was wrong, the equation should be
\[\cos^2(x)-\cos(x)-1=0\]

Answer 3

help solve please

Answer 4

probably need the quadratic formula
if you put \(u=\cos(x)\) then this is
\[u^2-u-1=0\]

Answer 5

do you know how to solve that one?

Answer 6

I don't understand any of it

Answer 7

ok lets go slow
and forget trig for a moment
have you ever solved a quadratic equation?

Answer 8

oh actually, before we begin, is it
\[\sin^2(x)+\cos(x)=0\] or is it
\[\sin(2x)+\cos(x)=0\]?