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Question Number 157324 by ZiYangLee last updated on 22/Oct/21
Find the general solution for the equation  cos 7x − cos 4x + cos x = 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}\:\mathrm{7}{x}\:−\:\mathrm{cos}\:\mathrm{4}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$
Commented by cortano last updated on 22/Oct/21
 cos 7x+cos x−cos 4x=0   2cos 4x cos 3x−cos 4x=0   cos 4x(2cos 3x−1)=0    { ((cos 4x=0)),((cos 3x=(1/2))) :}
$$\:\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\:\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\:\mathrm{cos}\:\mathrm{4}{x}\left(\mathrm{2cos}\:\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0}}\\{\mathrm{cos}\:\mathrm{3}{x}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$
Answered by gsk2684 last updated on 22/Oct/21
cos 7x+cos x−cos 4x=0  2cos (((7x+x)/2))cos (((7x−x)/2))−cos 4x=0  cos 4x(2cos 3x−1)=0  cos 4x=0 or cos 3x=(1/2)  4x=(2n+1)(π/2) or 3x=2nπ±(π/3) where n∈I  x=(2n+1)(π/8) or x=((2nπ)/3)±(π/9) where n∈I
$$\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\mathrm{2cos}\:\left(\frac{\mathrm{7}{x}+{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{7}{x}−{x}}{\mathrm{2}}\right)−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\left(\mathrm{2cos}\:\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0}\:{or}\:\mathrm{cos}\:\mathrm{3}{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:{or}\:\mathrm{3}{x}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}}\:{where}\:{n}\in{I} \\ $$$${x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{8}}\:{or}\:{x}=\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\pm\frac{\pi}{\mathrm{9}}\:{where}\:{n}\in{I} \\ $$

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