как решить:
5 cos X + 12 sin X = 13

[tex]cosx=cos(2* \frac{x}{2})=cos^{2}( \frac{x}{2})-sin^{2}( \frac{x}{2})[/tex]
[tex]sinx=sin(2* \frac{x}{2})=2*cos( \frac{x}{2})*sin( \frac{x}{2})[/tex]
[tex]13=13cos^{2}( \frac{x}{2})+13sin^{2}( \frac{x}{2})[/tex]

[tex]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[/tex]
[tex]-8cos^{2}( \frac{x}{2})-18sin^{2}( \frac{x}{2})+24*cos( \frac{x}{2})*sin( \frac{x}{2})=0[/tex]
[tex]4+9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})=0[/tex]
[tex]9tg^{2}( \frac{x}{2})-12tg( \frac{x}{2})+4=0[/tex]
Замена: [tex]tg( \frac{x}{2})=t[/tex]
[tex]9t^{2}-12t+4=0, D=0[/tex]
[tex]t= \frac{12}{18}=\frac{2}{3}[/tex]

[tex]tg( \frac{x}{2})=\frac{2}{3}[/tex]
[tex]\frac{x}{2}=arctg(\frac{2}{3})+ \pi k[/tex], k∈Z
[tex]x=2arctg(\frac{2}{3})+ 2\pi k[/tex], k∈Z