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Question Number 134165 by liberty last updated on 28/Feb/21

Solve (√(1+cos x)) + (√(1−cos x)) = 2sin x

$$\mathrm{Solve}\:\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\:+\:\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:=\:\mathrm{2sin}\:\mathrm{x} \\ $$

Answered by EDWIN88 last updated on 28/Feb/21

square both sides ⇔ 2+2(√(1−cos^2 x)) = 4sin^2 x ⇒ 2+2∣sin x∣ = 4sin^2 x ⇒2sin^2 x ± sin x −1 = 0 ⇒sin x = ((±1 ±3)/4) ∈ {−1,−(1/2),(1/2),2 } since the LHS is always nonnegative so the sine on the right must be nonnegative to. Also sin (x)=(1/2) turns out to be wrong a solution when we try to apply it. This leaves sin (x)=1 ⇒ x=2nπ +(π/2) or x = 2π(n+(1/4))

$$\mathrm{square}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\Leftrightarrow\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\:\mathrm{2}+\mathrm{2}\mid\mathrm{sin}\:\mathrm{x}\mid\:=\:\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\: \\ $$$$\Rightarrow\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\:\pm\:\mathrm{sin}\:\mathrm{x}\:−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{x}\:=\:\frac{\pm\mathrm{1}\:\pm\mathrm{3}}{\mathrm{4}}\:\in\:\left\{−\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\mathrm{2}\:\right\} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{LHS}\:\mathrm{is}\:\mathrm{always}\:\mathrm{nonnegative} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{must}\:\mathrm{be}\:\mathrm{nonnegative} \\ $$$$\mathrm{to}.\:\mathrm{Also}\:\mathrm{sin}\:\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{turns}\:\mathrm{out}\:\mathrm{to}\:\mathrm{be}\:\mathrm{wrong} \\ $$$$\mathrm{a}\:\mathrm{solution}\:\mathrm{when}\:\mathrm{we}\:\mathrm{try}\:\mathrm{to}\:\mathrm{apply}\:\mathrm{it}. \\ $$$$\mathrm{This}\:\mathrm{leaves}\:\mathrm{sin}\:\left(\mathrm{x}\right)=\mathrm{1}\:\Rightarrow\:\mathrm{x}=\mathrm{2n}\pi\:+\frac{\pi}{\mathrm{2}}\:\mathrm{or} \\ $$$$\mathrm{x}\:=\:\mathrm{2}\pi\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$

Commented by EDWIN88 last updated on 28/Feb/21

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