Sin^2x - 4sinxcosx + 3cos^2x=0

[latex] \frac{sin^2x}{cos^2x}- \frac{4sinxcosx}{cos^2x}+ \frac{3cos^2x}{cos^2x}= \frac{0}{cos^2x} \\ \\ tg^2x-4tgx+3=0 \\ \\ y=tgx \\ \\ y^2-4y+3=0 \\ D=16-12=4 \\ y_{1}= \frac{4-2}{2} =1 \\ \\ y_{2}= \frac{4+2}{2}=3 [/latex] При у=1 tgx=1 x=π/4 + πk, k∈Z При у=3 tgx=3 x=arctg3+πk, k∈Z Ответ: π/4+πk, k∈Z;             arctg3 +πk, k∈Z.