Advanced Functions
posted by Anonymous .
Consider the trig equation:
4sinxcos2x + 4cosxsin2x  1 = 0
in the Interval [0, 2 pi]
Either show that there is no solution to the equation in this domain, or determine the smallest possible solution.

4sinxcos2x + 4cosxsin2x  1 = 0
4(sinxcos2x + cosxsin2x) = 1
Did you recognize that the expression in my bracket matches with
sin(A+B) = sinAcosB +cosAsinB ?
so we have
4(sin(x+2x)) = 1
sin 3x = 1/4
3x = .25268 radians
x = .084227 radians
(you wanted the smallest possible solution)