Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 145654 by 7770 last updated on 07/Jul/21

4cosy−3secy=2tany Find y

$$\mathrm{4}{cosy}−\mathrm{3}{secy}=\mathrm{2}{tany} \\ $$$${Find}\:{y} \\ $$

Answered by john_santu last updated on 07/Jul/21

4cos y−(3/(cos y)) = ((2(√(1−cos^2 y)))/(cos y)) cos y≠ 0 ; let cos y = x ⇒4x−(3/x) = ((2(√(1−x^2 )))/x) ⇒4x^2 −3 = 2(√(1−x^2 )) let x^2 = d ⇒4d−3=2(√(1−d)) ⇒16d^2 −24d+9=4−4d ⇒16d^2 −20d+5=0 ⇒d = ((20 +(√(400−320)))/(32)) ⇒d=((20+4(√5))/(32)) = ((5+(√5))/8) ⇒cos^2 y = ((5+(√5))/8) ⇒cos y = ± (√((5+(√5))/8)) ⇒cos y= ± ((√(10+2(√5)))/4) y_1 = 2nπ ± arccos (((√(10+2(√5)))/4)) y_2 =2nπ ± arccos (−((√(10+2(√5)))/4))

$$\:\:\mathrm{4cos}\:{y}−\frac{\mathrm{3}}{\mathrm{cos}\:{y}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {y}}}{\mathrm{cos}\:{y}} \\ $$$$\:\mathrm{cos}\:{y}\neq\:\mathrm{0}\:;\:{let}\:\mathrm{cos}\:{y}\:=\:{x} \\ $$$$\Rightarrow\mathrm{4}{x}−\frac{\mathrm{3}}{{x}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} −\mathrm{3}\:=\:\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$$${let}\:{x}^{\mathrm{2}} \:=\:{d}\: \\ $$$$\Rightarrow\mathrm{4}{d}−\mathrm{3}=\mathrm{2}\sqrt{\mathrm{1}−{d}} \\ $$$$\Rightarrow\mathrm{16}{d}^{\mathrm{2}} −\mathrm{24}{d}+\mathrm{9}=\mathrm{4}−\mathrm{4}{d} \\ $$$$\Rightarrow\mathrm{16}{d}^{\mathrm{2}} −\mathrm{20}{d}+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{d}\:=\:\frac{\mathrm{20}\:+\sqrt{\mathrm{400}−\mathrm{320}}}{\mathrm{32}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{20}+\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{32}}\:=\:\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{2}} {y}\:=\:\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}\: \\ $$$$\Rightarrow\mathrm{cos}\:{y}\:=\:\pm\:\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}} \\ $$$$\Rightarrow\mathrm{cos}\:{y}=\:\pm\:\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$${y}_{\mathrm{1}} =\:\mathrm{2}{n}\pi\:\pm\:\mathrm{arccos}\:\left(\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\right) \\ $$$${y}_{\mathrm{2}} =\mathrm{2}{n}\pi\:\pm\:\mathrm{arccos}\:\left(−\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\right) \\ $$

Answered by MJS_new last updated on 07/Jul/21

4cos y −(3/(cos y))=2((sin y)/(cos y)) 4cos^2 y −3=2sin y 4(1−sin^2 y)−3=2sin y ⇔ sin^2 y +(1/2)sin y −(1/4)=0 sin y =−(1/4)±((√5)/4) ⇔ y=−((7π)/(10))+2nπ∨y=−((3π)/(10))+2nπ∨y=(π/(10))+2π∨y=((9π)/(10))+2π

$$\mathrm{4cos}\:{y}\:−\frac{\mathrm{3}}{\mathrm{cos}\:{y}}=\mathrm{2}\frac{\mathrm{sin}\:{y}}{\mathrm{cos}\:{y}} \\ $$$$\mathrm{4cos}^{\mathrm{2}} \:{y}\:−\mathrm{3}=\mathrm{2sin}\:{y} \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{y}\right)−\mathrm{3}=\mathrm{2sin}\:{y} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{y}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{y}\:−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{sin}\:{y}\:=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Leftrightarrow \\ $$$${y}=−\frac{\mathrm{7}\pi}{\mathrm{10}}+\mathrm{2}{n}\pi\vee{y}=−\frac{\mathrm{3}\pi}{\mathrm{10}}+\mathrm{2}{n}\pi\vee{y}=\frac{\pi}{\mathrm{10}}+\mathrm{2}\pi\vee{y}=\frac{\mathrm{9}\pi}{\mathrm{10}}+\mathrm{2}\pi \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com