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Question Number 18457 by Tinkutara last updated on 21/Jul/17
The number of solutions of the equation sin θ + cos θ = 1 + sin θ cos θ in the interval [0, 4π] is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interval}\:\left[\mathrm{0},\:\mathrm{4}\pi\right]\:\mathrm{is} \\ $$
Answered by mrW1 last updated on 21/Jul/17
sin θ + cos θ = 1 + sin θ cos θ 2(sin θ + cos θ) = 1 +sin^2 θ+cos^2 θ+2 sin θ cos θ 2(sin θ + cos θ) = 1 +( sin θ +cos θ)^2 1 −2(sin θ+cos θ)+( sin θ +cos θ)^2 =0 [(sin θ+cos θ)−1]^2 =0 sin θ+cos θ=1 sin θcos (π/4)+cos θsin (π/4)=((√2)/2) sin (θ+(π/4))=((√2)/2) θ+(π/4)=2kπ+(π/4),2kπ+((3π)/4) θ=2kπ,2kπ+(π/2) in [0,4π]: θ=0,(π/2),2π,((5π)/2),4π ⇒5 solutions
$$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\left(\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\right)\:=\:\mathrm{1}\:+\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\left(\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\right)\:=\:\mathrm{1}\:+\left(\:\mathrm{sin}\:\theta\:+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{1}\:−\mathrm{2}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)+\left(\:\mathrm{sin}\:\theta\:+\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left[\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)−\mathrm{1}\right]^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{sin}\:\theta\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{cos}\:\theta\mathrm{sin}\:\frac{\pi}{\mathrm{4}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\theta+\frac{\pi}{\mathrm{4}}=\mathrm{2k}\pi+\frac{\pi}{\mathrm{4}},\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\theta=\mathrm{2k}\pi,\mathrm{2k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{in}\:\left[\mathrm{0},\mathrm{4}\pi\right]: \\ $$$$\theta=\mathrm{0},\frac{\pi}{\mathrm{2}},\mathrm{2}\pi,\frac{\mathrm{5}\pi}{\mathrm{2}},\mathrm{4}\pi \\ $$$$\Rightarrow\mathrm{5}\:\mathrm{solutions} \\ $$
Commented by mrW1 last updated on 22/Jul/17
be careful with squaring! x=y ⇒ x^2 =y^2 but x^2 =y^2 ⇏ x=y alone since x^2 =y^2 ⇒ ∣x∣=∣y∣ i.e. x=±y in your case you have also solutions which fulfill following equation sin θ + cos θ = −(1 + sin θ cos θ) e.g. with θ=π sin π + cos π =−1 1+sin π cos π=1 1≠−1 but (1)^2 =(−1)^2
$$\mathrm{be}\:\mathrm{careful}\:\mathrm{with}\:\mathrm{squaring}! \\ $$$$\mathrm{x}=\mathrm{y}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{x}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} \:\nRightarrow\:\mathrm{x}=\mathrm{y}\:\mathrm{alone} \\ $$$$\mathrm{since}\:\mathrm{x}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} \:\Rightarrow\:\mid\mathrm{x}\mid=\mid\mathrm{y}\mid\:\mathrm{i}.\mathrm{e}.\:\mathrm{x}=\pm\mathrm{y} \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{your}\:\mathrm{case}\:\mathrm{you}\:\mathrm{have}\:\mathrm{also}\:\mathrm{solutions} \\ $$$$\mathrm{which}\:\mathrm{fulfill}\:\mathrm{following}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:−\left(\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right) \\ $$$$ \\ $$$$\mathrm{e}.\mathrm{g}.\:\mathrm{with}\:\theta=\pi \\ $$$$\mathrm{sin}\:\pi\:+\:\mathrm{cos}\:\pi\:=−\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{sin}\:\pi\:\mathrm{cos}\:\pi=\mathrm{1} \\ $$$$\mathrm{1}\neq−\mathrm{1}\:\mathrm{but}\:\left(\mathrm{1}\right)^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 22/Jul/17
Thanks Sir! But what is the mistake in my method? sin θ + cos θ = 1 + sin θ cos θ 1 + sin 2θ = 1 + sin^2 θ cos^2 θ + sin 2θ sin θ cos θ = 0 ⇒ sin 2θ = 0 θ = 0, (π/2), π, ((3π)/2), 2π, ((5π)/2), 3π, ((7π)/2), 4π
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{But}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in} \\ $$$$\mathrm{my}\:\mathrm{method}? \\ $$$$\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{1}\:+\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta\:+\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{0} \\ $$$$\theta\:=\:\mathrm{0},\:\frac{\pi}{\mathrm{2}},\:\pi,\:\frac{\mathrm{3}\pi}{\mathrm{2}},\:\mathrm{2}\pi,\:\frac{\mathrm{5}\pi}{\mathrm{2}},\:\mathrm{3}\pi,\:\frac{\mathrm{7}\pi}{\mathrm{2}},\:\mathrm{4}\pi \\ $$
Commented by Tinkutara last updated on 22/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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