Question Number 78946 by mathocean1 last updated on 21/Jan/20
$$\mathrm{solve} \\ $$$$\mathrm{cos}{x}−\sqrt{\mathrm{3}}{sinx}=\mathrm{1} \\ $$
Commented by msup trace by abdo last updated on 21/Jan/20
$${e}\:\Leftrightarrow\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{cosx}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}\right)=\mathrm{1}\:\Rightarrow \\ $$$${cos}\left(\frac{\pi}{\mathrm{3}}\right){cosx}−{sin}\left(\frac{\pi}{\mathrm{3}}\right){sinx}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{cos}\left({x}+\frac{\pi}{\mathrm{3}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right)\:\Rightarrow \\ $$$${x}+\frac{\pi}{\mathrm{3}}\:=\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:{or}\:{x}+\frac{\pi}{\mathrm{3}}=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:\left({kfrom}\:{Z}\right) \\ $$$$\Rightarrow{x}=\mathrm{3}{k}\pi\:{or}\:=\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi \\ $$
Commented by msup trace by abdo last updated on 21/Jan/20
$${error}\:{of}\:{typo}\:\:{x}=\mathrm{2}{k}\pi\:{or}\:{x}=\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi \\ $$
Answered by Mr. AR last updated on 31/Jan/20
$${cos}\left(\mathrm{90}°−{x}\right)−\sqrt{\mathrm{3}}\:{sinx}=\mathrm{1} \\ $$$${sin}\:{x}\:−\:\sqrt{\mathrm{3}}\:{sin}\:{x}=\mathrm{1} \\ $$$${sin}\:{x}\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\right)=\mathrm{1} \\ $$$${sin}\:{x}=\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{3}}}\:×\:\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}}\: \\ $$$${sin}\:{x}=\frac{\mathrm{1}+\mathrm{1}.\mathrm{732}}{\mathrm{1}−\mathrm{3}} \\ $$$${sin}\:{x}=\frac{\mathrm{2}.\mathrm{732}}{−\mathrm{2}} \\ $$$$\boldsymbol{{sin}}\:\boldsymbol{{x}}=−\mathrm{1}.\mathrm{366} \\ $$