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Question Number 17100 by virus last updated on 30/Jun/17

sec xcos 5x+1=0  find number of solution

$$\mathrm{sec}\:{x}\mathrm{cos}\:\mathrm{5}{x}+\mathrm{1}=\mathrm{0} \\ $$$${find}\:{number}\:{of}\:{solution} \\ $$

Answered by Tinkutara last updated on 01/Jul/17

cos 5x + cos x = 0  2 cos 3x cos 2x = 0  Either cos 3x = 0 or cos 2x = 0  ⇒ x = (2n + 1)(π/6) or (2k + 1)(π/4)  There should be infinite number of  solutions.

$$\mathrm{cos}\:\mathrm{5}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0} \\ $$$$\mathrm{Either}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{0}\:\mathrm{or}\:\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{6}}\:\mathrm{or}\:\left(\mathrm{2}{k}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{There}\:\mathrm{should}\:\mathrm{be}\:\mathrm{infinite}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{solutions}. \\ $$

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