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

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

$${Solve}\:{for}\:{real}\:{numbers}: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}{sin}\boldsymbol{{y}}−\mathrm{4}{cos}\boldsymbol{{y}}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$

Commented by iloveisrael 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

$$\Delta\geqslant\mathrm{0} \\ $$$$\mathrm{3sin}\:\mathrm{y}+\mathrm{4cosy}−\mathrm{6}\geqslant−\mathrm{1} \\ $$$$\mathrm{5}\:\:\mathrm{sin}\:\left(\mathrm{y}+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right)\geqslant\mathrm{5} \\ $$$$\mathrm{sin}\:\left(\mathrm{y}+\mathrm{tan}\:^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)=\:\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi \\ $$$$\Rightarrow\mathrm{y}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\mathrm{2k}\pi \\ $$

Commented by mathdanisur last updated on 06/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{Ser}} \\ $$

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)

$$\mathrm{3}{sin}\boldsymbol{{y}}−\mathrm{4}{cos}\boldsymbol{{y}}\:={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6} \\ $$$$\mathrm{5}\left({sin}\boldsymbol{{y}}\:\frac{\mathrm{3}}{\mathrm{5}}−{cos}\boldsymbol{{y}}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6} \\ $$$$\mathrm{5}\left({sin}\boldsymbol{{y}}\:\mathrm{cos}\:\alpha−{cos}\boldsymbol{{y}}\mathrm{sin}\:\alpha\right)\:=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5} \\ $$$${with}\:\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}},\:\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{5}\:\mathrm{sin}\:\left({y}−\alpha\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5} \\ $$$$\mathrm{5}\:\mathrm{sin}\:\left({y}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5} \\ $$$${LHS}\leqslant\mathrm{5} \\ $$$${RHS}\geqslant\mathrm{5} \\ $$$$\Rightarrow{LHS}={RHS}=\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\left({y}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}\right)=\mathrm{1} \\ $$$$\Rightarrow{y}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thank you Ser, cool

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{cool} \\ $$

Commented by peter frank last updated on 05/Aug/21

explanation 2nd and 3rd line

$${explanation}\:\mathrm{2}{nd}\:{and}\:\mathrm{3}{rd}\:{line} \\ $$

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

i added some lines more.

$${i}\:{added}\:{some}\:{lines}\:{more}. \\ $$

Commented by peter frank last updated on 05/Aug/21

thank you

$${thank}\:{you} \\ $$

Commented by Tawa11 last updated on 05/Aug/21

great

$$\mathrm{great} \\ $$

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

$${x}=\mathrm{1}\pm\sqrt{−\mathrm{5}+\mathrm{3sin}\:{y}\:+\mathrm{4cos}\:{y}} \\ $$$$−\mathrm{10}\leqslant−\mathrm{5}+\mathrm{3sin}\:{y}\:+\mathrm{4cos}\:{y}\:\leqslant\mathrm{0} \\ $$$$\Rightarrow \\ $$$$−\mathrm{5}+\mathrm{3sin}\:{y}\:+\mathrm{4cos}\:{y}\:=\mathrm{0}\:\Leftrightarrow\:{y}=\mathrm{2}{n}\pi+\mathrm{arctan}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\mathrm{solution}\:\mathrm{is} \\ $$$${x}=\mathrm{1}\wedge{y}=\mathrm{2}{n}\pi+\mathrm{arctan}\:\frac{\mathrm{3}}{\mathrm{4}}\forall{n}\in\mathbb{Z} \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thank you Ser, cool

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{cool} \\ $$

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