Giải phương trình:  1+sinx +cosx =2 cos (x/2- pi/4)

Trả lời

$1+\sin x+\cos x=2\cos \left(\frac{x}{2}-\frac{\pi }{4}\right)$

$\iff {\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}+{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}=2\left(\cos \frac{x}{2}.\cos \frac{\pi }{4}+\sin \frac{x}{2}.\sin \frac{\pi }{4}\right)$

$\iff 2{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}=2\left(\cos \frac{x}{2}.\frac{\sqrt{2}}{2}+\sin \frac{x}{2}.\frac{\sqrt{2}}{2}\right)$

$\iff 2\cos \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)=\sqrt{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)$

$\iff \left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)\left(\sqrt{2}\cos \frac{x}{2}-1\right)=0$

$\iff \left[\begin{array}{l}\sin \frac{x}{2}+\cos \frac{x}{2}=0\\ \sqrt{2}\cos \frac{x}{2}-1=0\end{array}\right.\iff \left[\begin{array}{l}\sqrt{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)=0\\ \cos \frac{x}{2}=\frac{\sqrt{2}}{2}\end{array}\right.$

$\iff \left[\begin{array}{l}\sin \left(\frac{x}{2}+\frac{\pi }{4}\right)=0\\ \cos \frac{x}{2}=\frac{\sqrt{2}}{2}\end{array}\right.\iff \left[\begin{array}{l}\frac{x}{2}+\frac{\pi }{4}=k\pi \\ \frac{x}{2}=\frac{\pi }{4}+k2\pi \\ \frac{x}{2}=-\frac{\pi }{4}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$

$\iff \left[\begin{array}{l}\frac{x}{2}=-\frac{\pi }{4}+k\pi \\ \frac{x}{2}=\frac{\pi }{4}+k2\pi \\ \frac{x}{2}=-\frac{\pi }{4}+k2\pi \end{array}\right.\left(k\in \mathbb{Z}\right)\iff \left[\begin{array}{l}x=-\frac{\pi }{2}+k2\pi \\ x=\frac{\pi }{2}+k4\pi \\ x=-\frac{\pi }{2}+k4\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$

$\iff x=\frac{\pi }{2}+k\pi \left(k\in \mathbb{Z}\right)$