cos(2x) csc^2(x) = 2cos(2x)
well, duh - divide by cos(2x), assuming cos(2x)≠0
csc^2(x) = 2
csc(x) = ±√2
x = all odd multiples of π/4
But, at those values of x, cos(2x)=0. So that set of roots is verboten.
Naturally the equation is true if cos(2x) = 0, meaning 2x is an odd multiple of π/2. Or, again, x is an odd multiple of π/4.
http://www.wolframalpha.com/input/?i=cos%282x%29+csc%5E2%28x%29+%3D+2cos%282x%29
identify all solutions in the interval [0,2pi]: cos (2x) csc^2x=2cos(2x)
I have no clue how to solve.
2 answers
thanks