don't know how to use double or half angle formula to solve the equation:
sin2x+cosx=0
in the interval [0, 2pi) and the answers are pi/2, 7pi/6, 3pi/2, 11pi/6
Substitute $\displaystyle \sin (2x) = 2 \sin x \cos x$.
Now factorise the resulting expression. Now apply the null factor law to get two simple trig equations that you need to solve over the given interval:
$\displaystyle \cos x = 0$
$\displaystyle 2 \sin (x) + 1 = 0$