Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com