Question

Rezolvati ecuatia 
$sin2x = 2 sin^{2}x$

Answer 1

Formula: sin(2x)=2*sin(x)*cos(x), oricare ar fi x numar real.

Deci sin(2x)=2(sin(x))^2 <=>2sin(x)cos(x)=2(sin(x))^2 <=> sin(x)cos(x)=(sin(x))^2 <=> sin(x)=0 sau cos(x)=sin(x) <=> sin(x)=0 sau tg(x)=1 <=> x apartine {k*pi | k apartine Z} sau x apartine{pi/4+k*pi | k apartine Z}.

In concluzie, multimea solutilor ecuatiei din enunt este S={k*pi | k apartine Z}U{pi/4 +k*pi | k apartine Z}.