Đặt sin x = a; cos x = b
Ta có hpt:
\(\left\{{}\begin{matrix}a+b=\sqrt{2}\\a^2+b^2=1\end{matrix}\right.\Rightarrow ab=\frac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}=\frac{2-1}{2}=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\sqrt{2}\\ab=\frac{1}{2}\end{matrix}\right.\Rightarrow a\cdot\left(\sqrt{2}-a\right)=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{2}a-a^2=\frac{1}{2}\Leftrightarrow a^2-\sqrt{2}a+\frac{1}{2}=0\)
\(\Leftrightarrow a^2-2\cdot a\cdot\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}\right)^2=0\)
\(\Leftrightarrow\left(a-\frac{\sqrt{2}}{2}\right)^2=0\Leftrightarrow a=\frac{\sqrt{2}}{2}\Rightarrow b=\frac{\sqrt{2}}{2}\)
Suy ra: \(\sin x=\cos x=\frac{\sqrt{2}}{2}\Rightarrow x=45^0\)
