Question

\(1+\sqrt{2}Sin\left(X+\frac{Π}{4}\right)+sin2x+cos2x=0\)

Answer 1

phương trình tương đương:

sin²x+cos²x+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))+2sinx.cosx+cos²x-sin²x=0

<=> 2cos²x+2sinx.cosx+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0

<=> 2cosx(sinx+cosx)+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0

<=>(2cosx+1).\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0

<=>\(\left[\begin{array}{nghiempt}sin\left(x+\frac{\pi}{4}\right)=0\\2cosx+1=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=k\pi-\frac{\pi}{4}\\x=\pm\frac{1}{2}+k2\pi\end{array}\right.\)với k\(\in\)Z

pt có 2 nghiệm như trên