x∈[0; 2π], [tex] sin^{2}x=sinx [/tex]⇒2[tex] sin^{2}x-sinx=0 [/tex] de unde:
sinx(2sinx-1)=0, egalam fiecare factor cu 0:
sinx=1 ⇒x=0 ; x=π si x=2π ; apoi 2sinx=1 ⇒ sinx=1/2, care are solutiile x=π/6 si
x=π-π/6=(5π)/6. deci solutia est : x∈{0; π/6; 5π/6; π,2π}
