Răspuns:
d. tgx=tg2x -->tgx- , deci tgx(1-)=0. din tgx=0 --.> x∈{kπ / k∈Z} . Egaland paranteza cu 0 si eliminand numitorul ( conditia 1-(tgx)^2≠0 ) obtinem ecuatia; 1--2=0, adica imposibil, deci avem numai solutia gasita anterior.
f. sin2x = cos3x
sin2x = sin(90-3x)
So, 2x=90-3x
5x=90
x=18
Answer is 18
e.ctg 2x=(ctg²x-1)/2 ctg x=(9-1)/2.3=8/6=4/3
