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

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