Как решить: 5 cos X + 12 sin X = 13

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

$5cos^{2}( \frac{x}{2})-5sin^{2}( \frac{x}{2})+12*2*cos( \frac{x}{2})*sin( \frac{x}{2})-13cos^{2}( \frac{x}{2})-13sin^{2}( \frac{x}{2})=0$
$-8cos^{2}( \frac{x}{2})-18sin^{2}( \frac{x}{2})+24*cos( \frac{x}{2})*sin( \frac{x}{2})=0$
$4+9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})=0$
$9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})+4=0$
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
$x=2arctg(\frac{2}{3})+ 2\pi k$ , k∈Z