Consider the trig equation:

Question

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.

Answer 1

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)