Question

2sin^2x+9sinxcosx+10cos^2x=0

Answer 1

2sin²x +9sinxcosx +10cos²x =0  * * *  cos²x ≠0 * * *
2tq²x +9tqx +10 =0 ;  * * * замена  t =tqx  * * *
2t² +9t +10 =0  ;
D =9² -4*2*10 =1² ;
t ₁= (-9 -1)/(2*2) = -2,5 ⇒ tqx₁ = -2,5⇒ x₁ = -arctq(2,5) +π*n , n∈Z.
t₂ = (-9 +1)/(2*2)  = -2 ⇒ tqx₂ = -2⇒ x₂ = -arctq2 +π*n , n∈Z.

ответ : - arctq(2,5) +π*n    ;   - arctq2 +π*n , n∈Z.