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Question Number 156482 by MathSh last updated on 11/Oct/21

Solve for real numbers: 3 + sin(2x) = 4sin(x + (π/4))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{3}\:+\:\mathrm{sin}\left(\mathrm{2x}\right)\:=\:\mathrm{4sin}\left(\mathrm{x}\:+\:\frac{\pi}{\mathrm{4}}\right) \\ $$

Answered by mr W last updated on 11/Oct/21

3+sin (2x)=2(√2)(sin x+cos x) 9+6sin (2x)+sin^2 (2x)=8(1+sin (2x)) sin^2 (2x)−2 sin (2x)+1=0 ⇒sin (2x)=1 ...(i) 3+1 = 4sin(x + (π/4)) ⇒sin (x+(π/4))=1 ..(ii) x+(π/4)=2kπ+(π/2) ⇒x=2kπ+(π/4) it fulfills both (i) and (ii).

$$\mathrm{3}+\mathrm{sin}\:\left(\mathrm{2}{x}\right)=\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{9}+\mathrm{6sin}\:\left(\mathrm{2}{x}\right)+\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2}{x}\right)=\mathrm{8}\left(\mathrm{1}+\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2}{x}\right)−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{2}{x}\right)=\mathrm{1}\:\:\:...\left({i}\right) \\ $$$$\mathrm{3}+\mathrm{1}\:=\:\mathrm{4sin}\left(\mathrm{x}\:+\:\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{1}\:\:\:..\left({ii}\right) \\ $$$${x}+\frac{\pi}{\mathrm{4}}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{4}} \\ $$$${it}\:{fulfills}\:{both}\:\left({i}\right)\:{and}\:\left({ii}\right). \\ $$

Commented by MathSh last updated on 11/Oct/21

Very nice solution dear Ser, thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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