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Question Number 18456 by Tinkutara last updated on 21/Jul/17

The complete solution of the equation sin 2x − 12(sin x − cos x) + 12 = 0 is given by (1) x = 2nπ + (π/2), (2n − 1)(π/4), n ∈ Z (2) x = nπ + (π/2), (2n + 1)π, n ∈ Z (3) x = 2nπ + (π/2), (2n + 1)π, n ∈ Z (4) x = nπ + (π/2), (2n − 1)π, n ∈ Z

$$\mathrm{The}\:\mathrm{complete}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:−\:\mathrm{12}\left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)\:+\:\mathrm{12}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)\frac{\pi}{\mathrm{4}},\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{4}\right)\:{x}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)\pi,\:{n}\:\in\:{Z} \\ $$

Answered by Tinkutara last updated on 30/Jul/17

sin 2x = 1 − (sin x − cos x)^2 = 1 − t^2 t^2 + 12t − 13 = 0 ⇒ cos x − sin x = −1 ⇒ cos (x + (π/4)) = −1 ⇒ x = 2nπ + (π/2), 2nπ − π

$$\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{1}\:−\:\left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} \:=\:\mathrm{1}\:−\:{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} \:+\:\mathrm{12}{t}\:−\:\mathrm{13}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{cos}\:{x}\:−\:\mathrm{sin}\:{x}\:=\:−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left({x}\:+\:\frac{\pi}{\mathrm{4}}\right)\:=\:−\mathrm{1} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{2}},\:\mathrm{2}{n}\pi\:−\:\pi \\ $$

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