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Question Number 134872 by bemath last updated on 08/Mar/21

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$$−−−−−−−−−−−−− \\ $$ Trygonometri What is the value of cosA if cosA+3sinA=2?\\n

Answered by EDWIN88 last updated on 08/Mar/21

We have sin A = ± (√(1−cos ^2 A)) then 3sin A = 2−cos A , square to get 9sin ^2 A = (2−cos A)^2 9−9cos ^2 A = cos ^2 A−4cos A+4 10cos ^2 A−4cos A−5= 0 cos A = ((4± (√(16+4.50)))/(20)) = ((4 ± 2(√(54)))/(20)) cos A = ((2±3(√6))/(10))

$$\mathrm{We}\:\mathrm{have}\:\mathrm{sin}\:\mathrm{A}\:=\:\pm\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}} \\ $$ $$\mathrm{then}\:\mathrm{3sin}\:\mathrm{A}\:=\:\mathrm{2}−\mathrm{cos}\:\mathrm{A}\:,\:\mathrm{square}\:\mathrm{to}\:\mathrm{get} \\ $$ $$\mathrm{9sin}\:^{\mathrm{2}} \mathrm{A}\:=\:\left(\mathrm{2}−\mathrm{cos}\:\mathrm{A}\right)^{\mathrm{2}} \\ $$ $$\mathrm{9}−\mathrm{9cos}\:^{\mathrm{2}} \mathrm{A}\:=\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}−\mathrm{4cos}\:\mathrm{A}+\mathrm{4}\: \\ $$ $$\mathrm{10cos}\:^{\mathrm{2}} \mathrm{A}−\mathrm{4cos}\:\mathrm{A}−\mathrm{5}=\:\mathrm{0}\: \\ $$ $$\:\mathrm{cos}\:\mathrm{A}\:=\:\frac{\mathrm{4}\pm\:\sqrt{\mathrm{16}+\mathrm{4}.\mathrm{50}}}{\mathrm{20}}\:=\:\frac{\mathrm{4}\:\pm\:\mathrm{2}\sqrt{\mathrm{54}}}{\mathrm{20}} \\ $$ $$\:\mathrm{cos}\:\mathrm{A}\:=\:\frac{\mathrm{2}\pm\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{10}}\: \\ $$

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