Cho sinx+cosx=1/2 tính giá trị vủa x

$\begin{array}{l} \sin x + \cos x = \frac{1}{2} \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{2\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \arcsin \frac{1}{{2\sqrt 2 }} + k2\pi \\ x + \frac{\pi }{4} = \pi - \arcsin \frac{1}{{2\sqrt 2 }} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{4} + \arcsin \frac{1}{{2\sqrt 2 }} + k2\pi \\ x = \frac{{3\pi }}{4} - \arcsin \frac{1}{{2\sqrt 2 }} + k2\pi \end{array} \right. \end{array}$

$\sin x+\cos x=\dfrac{1}{2}$

$\Leftrightarrow \sqrt2\sin\Big(x+\dfrac{\pi}{4}\Big)=\dfrac{1}{2}$

$\Leftrightarrow \sin\Big(x+\dfrac{\pi}{4}\Big)=\dfrac{1}{2\sqrt2}$

$\Leftrightarrow x=\dfrac{-\pi}{4}+\arcsin\dfrac{1}{2\sqrt2}+k2\pi$

hoặc $x=\dfrac{3\pi}{4}-\arcsin\dfrac{1}{2\sqrt2}+k2\pi$