2ctgx-3tgx+5=0 Решите пожалуйста!

x\neq\frac{\pi k}{2}, k\in Z\\ x\in(\frac{\pi k}{2};\frac{\pi+\pi k}{2});\\ tgx=t;\ \ ctgx=\frac{1}{tgx}=\frac1t;\\ \frac2t-3t+5=0;\\ t\neq0;\pm\infty;\\ -3t^2+5t+2=0;\\ 3t^2-5t-2=0;\\ D=b^2-4\cdot a\cdot c=(-5)^2-4\cdot3\cdot(-2)=25+24=49=(\pm7);\\ " alt="2ctgx-3tgx+5=0;\\ D(f): \left \{ {{x\neq \pi k} \atop {x\neq\frac\pi2+\pi l}} \right. \ k,l\in Z==>x\neq\frac{\pi k}{2}, k\in Z\\ x\in(\frac{\pi k}{2};\frac{\pi+\pi k}{2});\\ tgx=t;\ \ ctgx=\frac{1}{tgx}=\frac1t;\\ \frac2t-3t+5=0;\\ t\neq0;\pm\infty;\\ -3t^2+5t+2=0;\\ 3t^2-5t-2=0;\\ D=b^2-4\cdot a\cdot c=(-5)^2-4\cdot3\cdot(-2)=25+24=49=(\pm7);\\ " align="absmiddle" class="latex-formula">
$t_1=\frac{-b-\sqrt D}{2\cdot a}=\frac{5-7}{2\cdot3}=\frac{-2}{6}=-\frac13;\ x_1=-arctg\frac13+\pi n\\ t_2=\frac{-b+\sqrt D}{2\cdot a}=\frac{5+7}{2\cdot3}=\frac{12}{6}=2;x_2=arctg2+\pi n\\$