Question

1) sin^2x-2sinxcosx=0

Answer 1

$1 - sin^{2}x - 2sinxcosx = 0 \cos^{2}x - 2 sinxcosx=0 \cosx(cosx - 2sinx)=0 \a) cosx = 0 \x = pi/2 + pi n \b) cosx-2sinx=0 | : cosx \1-2tgx=0 \tgx=frac{1}{2} \x=arctgfrac{1}{2} + pi n$

n ∈ Z

 

Ответ: х₁ = π/2 + πn, n∈Z; x₂ = arctg½ + πn, n∈Z