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Question Number 83811 by Power last updated on 06/Mar/20

Commented by Power last updated on 06/Mar/20

thanks

$$\mathrm{thanks} \\ $$

Commented by john santu last updated on 06/Mar/20

sin (41+x) = sin (x+19) + sin (x−19) ⇒ sin (41+x)−sin (x−19) = sin (x+19) 2cos (x+11) sin 30 = sin (x+19) cos (x+11) = sin (x+19) cos (x+11) = cos (90−x−19) ⇒ x+11 = 71−x +2kπ 2x = 60 + 2kπ ⇒ x = 30 + kπ

$$\mathrm{sin}\:\left(\mathrm{41}+\mathrm{x}\right)\:=\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{19}\right)\:+\:\mathrm{sin}\:\left(\mathrm{x}−\mathrm{19}\right) \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\mathrm{41}+\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}−\mathrm{19}\right)\:=\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{19}\right) \\ $$$$\mathrm{2cos}\:\left(\mathrm{x}+\mathrm{11}\right)\:\mathrm{sin}\:\mathrm{30}\:=\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{19}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{x}+\mathrm{11}\right)\:=\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{19}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{x}+\mathrm{11}\right)\:=\:\mathrm{cos}\:\left(\mathrm{90}−\mathrm{x}−\mathrm{19}\right) \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{11}\:=\:\mathrm{71}−\mathrm{x}\:+\mathrm{2k}\pi \\ $$$$\mathrm{2x}\:=\:\mathrm{60}\:+\:\mathrm{2k}\pi\:\Rightarrow\:\mathrm{x}\:=\:\mathrm{30}\:+\:\mathrm{k}\pi \\ $$$$ \\ $$

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