điều kiện : \(-1\le sinx;cosx\le1\)
ta có : \(sinx+cosx=\dfrac{1}{2}\Leftrightarrow cosx=\dfrac{1}{2}-sinx\)
ta lại có : \(sin^2x+cos^2x=1\Leftrightarrow sin^2x+\left(\dfrac{1}{2}-sinx\right)^2=1\)
\(\Leftrightarrow sin^2x+sin^2x-sinx+\dfrac{1}{4}=1\Leftrightarrow2sin^2x-sinx-\dfrac{3}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1+\sqrt{7}}{4}\\sinx=\dfrac{1-\sqrt{7}}{4}\end{matrix}\right.\)
ta có : (+) \(sinx=\dfrac{1+\sqrt{7}}{4}\Rightarrow cosx=\dfrac{1-\sqrt{7}}{4}\left(tmđk\right)\)
(+) \(sinx=\dfrac{1-\sqrt{7}}{4}\Rightarrow cos=\dfrac{1+\sqrt{7}}{4}\left(tmđk\right)\)
vậy ..........................................................................................................................
