Question

5sin^2x+3sinxcosx-3cos^2x=2​

Answer 1

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

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

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

$left { {{cosxneq 0} atop {tgx=frac{-3бsqrt{69} }{6} } right.$

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