Find all solutions in the interval [0, 2pi)
2sin^2 x = 2 + cox x
I kno that most of the solutions are gonna be 1st quadrant solutions,
sin is positive in the 1st and 2nd and cos is negative in the 4th quadrant
Im coming up with cos(x)=0 and cos(x)=-1, but the interval is [0, 2pi) with those answers it could only be 1 solution but it asked me to find solutions yo i suck at this class, i might have tell my guidance counselor to switch me out.....sorry if im buiggin the mess out of you
from the graph of cos(x) we know that $\displaystyle \cos x = 0 \implies x = \frac {\pi}2 + k \pi$ for $\displaystyle k \in \mathbb{Z}$
from the table i gave you in one of your other posts and your knowledge of reference angles, we know $\displaystyle \cos x = - \frac 12 \implies x = \frac {2 \pi}3 + k \pi \mbox{ or } \frac {4 \pi}3 + n \pi$ for $\displaystyle n,k \in \mathbb{Z}$
(cosine is negative in the second and third quadrants)