Как решить: 5 cos X + 12 sin X = 13

[latex]cosx=cos(2* \frac{x}{2})=cos^{2}( \frac{x}{2})-sin^{2}( \frac{x}{2})[/latex] [latex]sinx=sin(2* \frac{x}{2})=2*cos( \frac{x}{2})*sin( \frac{x}{2})[/latex] [latex]13=13cos^{2}( \frac{x}{2})+13sin^{2}( \frac{x}{2})[/latex] [latex]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[/latex] [latex]-8cos^{2}( \frac{x}{2})-18sin^{2}( \frac{x}{2})+24*cos( \frac{x}{2})*sin( \frac{x}{2})=0[/latex] [latex]4+9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})=0[/latex] [latex]9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})+4=0[/latex] Замена: [latex]tg( \frac{x}{2})=t[/latex] [latex]9t^{2}-12t+4=0, D=0[/latex] [latex]t= \frac{12}{18}=\frac{2}{3}[/latex] [latex]tg( \frac{x}{2})=\frac{2}{3}[/latex] [latex]\frac{x}{2}=arctg(\frac{2}{3})+ \pi k[/latex], k∈Z [latex]x=2arctg(\frac{2}{3})+ 2\pi k[/latex], k∈Z