Question

cos^2x+sin2x-3sin^2x=0 решите уравнение

Answer 1

$frac{cos^2x}{cos^2x}+ frac{2sinxcosx}{cos^2x}- frac{3sin^2x}{cos^2x}= frac{0}{cos^2x}$

1+2tgx-3tg²x=0
y=tgx

1+2y-3y²=0
3y²-2y-1=0
D=4+12=16
y₁=(2-4)/6=-2/6=-1/3
y₂=(2+4)/6=1

При у= -1/3
tgx= -1/3
x= -arctg(1/3)+πk, k∈Z

При у=1
tgx=1
x=π/4 + πk, k∈Z