Question

как решить:
5 cos X + 12 sin X = 13

Answer 1

$cosx=cos(2* frac{x}{2})=cos^{2}( frac{x}{2})-sin^{2}( frac{x}{2})$
$sinx=sin(2* frac{x}{2})=2*cos( frac{x}{2})*sin( frac{x}{2})$
$13=13cos^{2}( frac{x}{2})+13sin^{2}( frac{x}{2})$

$5cos^{2}( frac{x}{2})-5sin^{2}( frac{x}{2})+12*2*cos( frac{x}{2})*sin( frac{x}{2})-13cos^{2}( frac{x}{2})-13sin^{2}( frac{x}{2})=0$
$-8cos^{2}( frac{x}{2})-18sin^{2}( frac{x}{2})+24*cos( frac{x}{2})*sin( frac{x}{2})=0$
$4+9tg^{2}( frac{x}{2})-12tg( frac{x}{2})=0$
$9tg^{2}( frac{x}{2})-12tg( frac{x}{2})+4=0$
Замена: $tg( frac{x}{2})=t$
$9t^{2}-12t+4=0, D=0$
$t= frac{12}{18}=frac{2}{3}$

$tg( frac{x}{2})=frac{2}{3}$
$frac{x}{2}=arctg(frac{2}{3})+ pi k$, k∈Z
$x=2arctg(frac{2}{3})+ 2pi k$, k∈Z