Question

5sin^2x+3sinxcosx-3cos^2x=2​

Answer 1

$5sin^2x+3sinx*cosx-3cos^2x-2(cos^2x+sin^2x)=0;\\3sin^2x+3sinx*cosx-5cos^2x=0;\\\left[\begin{array}{ccc}\left \{ {{cos^2x=0} \atop {3sin^2x+3sinx*0-5*0=0}} \right. \\\left \{ {{cos^2x\neq 0} \atop {cos^2x(3tg^2x+3tgx-5)=0}} \right. \\\end{array}$

$\left[\begin{array}{ccc}\left \{ {{cos^2x=0} \atop {sin^2x=0}}; cos^2x+sin^2x\neq 0=>net+resheniiy\right. \\\left \{ {{cos^2x\neq 0} \atop {3tg^2x+3tgx-5=0}} \right. \\\end{array}$

tgx=a; D=9+20*3=69

$\left \{ {{cosx\neq 0} \atop {tgx=\frac{-3б\sqrt{69} }{6} } \right.$

Ответ: x={arctg((-3±√69)/6)+pi*n}, n∈Z.