Question

13sin^2(x)+84sin2x-13cos^2(x)+1=2sin18*cos18/cos54

Answer 1

$\displaystyle 13sin^2x+84sin(2x)-13cos^2x+1= \frac{2sin18*cos18}{cos54}$

$\displaystyle 13sin^2x+168sinx*cosx-13cos^2x+1= \frac{sin 36}{cos54}$

$\displaystyle 13sin^2x+168sinx*cosx-13cos^2x= \frac{cos(90-36)}{cos54}-1$

$\displaystyle 13sin^2x+168sinx*cosx-13cos^2x=0$

разделим на cos²x≠0

$\displaystyle 13tg^2x+168tgx-13=0 D=28224+676=28900=170^2$

$\displaystyle tg x=(-168+170)/26=1/13$

$\displaystyle tg x=(-168-170)/26=-13$

$\displaystyle x=arctg (-13)+ \pi n, n\in Z$

$\displaystyle x=arctg(1/13)+ \pi n, n\in Z$