Question Number 149478 by mathdanisur last updated on 05/Aug/21

Solve for real numbers: x^2 −2x−3siny−4cosy + 6 = 0

Commented byiloveisrael last updated on 06/Aug/21

Δ≥0 3sin y+4cosy−6≥−1 5 sin (y+tan^(−1) ((4/3)))≥5 sin (y+tan ^(−1) ((4/3)))=1 ⇒y+tan^(−1) ((4/3))= (π/2)+2kπ ⇒y=(π/2)−tan^(−1) ((4/3))+2kπ

Commented bymathdanisur last updated on 06/Aug/21

Thankyou Ser

Answered by mr W last updated on 05/Aug/21

3siny−4cosy =x^2 −2x+6 5(siny (3/5)−cosy (4/5)) =x^2 −2x+6 5(siny cos α−cosysin α) =(x−1)^2 +5 with cos α=(3/5), sin α=(4/5), tan α=(4/3) 5 sin (y−α)=(x−1)^2 +5 5 sin (y−tan^(−1) (4/3))=(x−1)^2 +5 LHS≤5 RHS≥5 ⇒LHS=RHS=5 ⇒x=1 ⇒sin (y−tan^(−1) (4/3))=1 ⇒y−tan^(−1) (4/3)=2kπ+(π/2) ⇒y=2kπ+(π/2)+tan^(−1) (4/3)

Commented bymathdanisur last updated on 05/Aug/21

Thank you Ser, cool

Commented bypeter frank last updated on 05/Aug/21

explanation 2nd and 3rd line

Commented bymr W last updated on 05/Aug/21

i added some lines more.

Commented bypeter frank last updated on 05/Aug/21

thank you

Commented byTawa11 last updated on 05/Aug/21

great

Answered by MJS_new last updated on 05/Aug/21

x=1±(√(−5+3sin y +4cos y)) −10≤−5+3sin y +4cos y ≤0 ⇒ −5+3sin y +4cos y =0 ⇔ y=2nπ+arctan (3/4) ⇒ solution is x=1∧y=2nπ+arctan (3/4)∀n∈Z

Commented bymathdanisur last updated on 05/Aug/21

Thank you Ser, cool