Question

3sinx-8cosx=3
Пожалуйста)

Answer 1

Ответ:

Объяснение:

$3sin(x)-8cos(x)=3<=>3\frac{2tg(\frac{x}{2} )}{1+tg^2(\frac{x}{2} )} -8\frac{1-tg^2(\frac{x}{2} )}{1+tg^2(\frac{x}{2} )} =3\\tg(\frac{x}{2} )=t=>\frac{6t-(8-8t^2)-3(1+t^2)}{1+t^2} =0<=>5t^2+6t-11=0\\t=1=>tg(\frac{x}{2} )=1=>x=\frac{\pi}{2}+2\pi k\\ t=-\frac{11}{5} =>tg(\frac{x}{2} )=-\frac{11}{5}=>x=-2arctg(\frac{11}{5} )+2\pi k$