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Question Number 189952 by SAMIRA last updated on 25/Mar/23

$$\sqrt{2}\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)=-\sqrt{2}$$

Answered by cortano12 last updated on 25/Mar/23

$$\sqrt{3}(\frac{\sqrt{2}}{\sqrt{3}}\sin \mathrm{x}+\frac{1}{\sqrt{3}}\cos \mathrm{x})=-\sqrt{2}$$$$\sqrt{3}(\sin \mathrm{x}\cos \theta +\cos \mathrm{x}\sin \theta )=-\sqrt{2}$$$$\sin (\mathrm{x}+\theta )=-\frac{\sqrt{2}}{\sqrt{3}}=\sin (180°+\alpha )$$$$\mathrm{where}\{\begin{array}{l}\theta =35°\\ \alpha =55°\end{array}$$

Answered by mehdee42 last updated on 25/Mar/23

$$lety=tan\frac{x}{2}\Rightarrow \sqrt{2}\times \frac{2y}{1+y^2}+\frac{1-y^2}{1+y^2}=-\sqrt{2}$$$$(1-\sqrt{2})y^2-2\sqrt{2}y-\sqrt{2}-1=0$$$$y=-{(\sqrt{2}+1)}^2=tan\alpha \Rightarrow x=2k\pi +2\alpha$$$$y=-1=tan(-\frac{\pi }{4})\Rightarrow x=2k\pi -\frac{\pi }{2}$$

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