Question

solve the equation: 2cos^2(x) + 3cos(x) +cos^2(x) + sin^2(x) = 0 where 0 <= x < 2pi

Answer 1

\[
2\cos^2(x) + 3\cos(x) +\cos^2(x) + \sin^2(x) =0\\
2\cos^2(x) + 3\cos(x) +1 =0\\
(\cos(x) +1)( 2\cos(x) +1)=0

\]
Can you finish it now?

Answer 2

\[
\cos(x) =-1\implies x = \pi \text{ Since we need answers between 0 and } 2\pi
\]

Answer 3

\[
\cos(x)=-\frac 1 2\implies x=\frac{2 \pi }3 \text {or } x=\frac{4 \pi }3 \text {Since we need answers between 0 and} 2\pi
\]

Answer 4

whoa never thought to factor XD thank you so much!

Answer 5

YW