Use the equation
sin(2x) = 2 sin x cos x
so that
6 sinx cosx -2 cosx = 0
cosx*(1 - 3sinx) = 0
(I divided out -2)
So the equation is satisfied wherever cosx = 0 or sinx = 1/3
x = pi/2 and 0.3398 radians are two solutions. There are others in other quadrants.
3sin(2x)-2cos(x)=0
1 answer