Giải phương trình :sinx+sin2x=cosx+cos2x

`\qquad sinx+sin2x=cosx+cos2x`

`<=>sinx-cosx=cos2x-sin2x`

`<=>\sqrt{2}/2 sinx-\sqrt{2}/2 cosx=\sqrt{2}/2 cos2x-\sqrt{2}/2 sin2x`

`<=>sinx.cos\  π/4-cosx sin\ π/4=sin\ π/4 cos2x-cos\ π/4sin2x`

`<=>sin(x-π/4)=sin(π/4-2x)`

$⇔\left[\begin{array}{l}x-\dfrac{π}{4}=\dfrac{π}{4}-2x+k2π\\x-\dfrac{π}{4}=π-(\dfrac{π}{4}-2x)+k2π\end{array}\right. \ (k\in Z)$

$⇔\left[\begin{array}{l}3x=\dfrac{π}{2}+k2π\\-x=π+k2π\end{array}\right.\ (k\in Z)$

$⇔\left[\begin{array}{l}x=\dfrac{π}{6}+\dfrac{k2π}{3}\\x=-π-k2π\end{array}\right.\ (k\in Z)$