Question Number 189952 by SAMIRA last updated on 25/Mar/23
$$\sqrt{\mathrm{2}}\:\mathrm{sin}\left(\mathrm{x}\right)\:+\mathrm{cos}\left(\mathrm{x}\right)=\:−\sqrt{\mathrm{2}}\: \\ $$
Answered by cortano12 last updated on 25/Mar/23
$$\:\sqrt{\mathrm{3}}\:\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\:\mathrm{x}\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\sqrt{\mathrm{3}}\:\left(\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\theta+\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\theta\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\:\mathrm{sin}\:\left(\mathrm{x}+\theta\right)=−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:=\mathrm{sin}\:\left(\mathrm{180}°+\alpha\right) \\ $$$$\:\mathrm{where}\:\begin{cases}{\theta=\mathrm{35}°}\\{\alpha=\mathrm{55}°}\end{cases} \\ $$
Answered by mehdee42 last updated on 25/Mar/23
$${let}\:\:{y}={tan}\frac{{x}}{\mathrm{2}}\Rightarrow\sqrt{\mathrm{2}}×\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }+\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }=−\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{y}−\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$${y}=−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} ={tan}\alpha\Rightarrow{x}=\mathrm{2}{k}\pi+\mathrm{2}\alpha \\ $$$${y}=−\mathrm{1}={tan}\left(−\frac{\pi}{\mathrm{4}}\right)\Rightarrow{x}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}} \\ $$