Question

10cos2x+8=tgx помогите решить

Answer 1

$10\cos 2x+8=tg x \\ \\ 10\cdot \frac{1-tg^2x}{1+tg^2x} +8=tgx$

Пусть tg²x = t (t≥0)
$10 \cdot \frac{1-t^2}{1+t^2} +8=t|\cdot (1+t^2) \\ \\ 10-10t^2+8+8t^2=t+t^3\\ -2t^2+18=t+t^2 \\ t^3+2t^2+t-18=0 \\ t^3-2t^2+4t^2-8t+9t-18=0 \\ t^2(t-2)+4t(t-2)+9(t-2)=0 \\ (t-2)(t^2+4t+9)=0 \\ t=2$

Обратная замена
$tg^2x=2 \\ tgx=\pm \sqrt{2} \\ x=\pm arctg(\sqrt{2})+ \pi n,n \in Z$