Toán Giải phương trình :sinx+sin2x=cosx+cos2x 18/09/2021 By Eliza 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)` $⇔\left[\begin{array}{l}x-\dfrac{π}{4}=\dfrac{π}{4}-2x+k2π\\x-\dfrac{π}{4}=π-(\dfrac{π}{4}-2x)+k2π\end{array}\right. \ (k\in Z)$ $⇔\left[\begin{array}{l}3x=\dfrac{π}{2}+k2π\\-x=π+k2π\end{array}\right.\ (k\in Z)$ $⇔\left[\begin{array}{l}x=\dfrac{π}{6}+\dfrac{k2π}{3}\\x=-π-k2π\end{array}\right.\ (k\in Z)$ Trả lời
`\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)`
$⇔\left[\begin{array}{l}x-\dfrac{π}{4}=\dfrac{π}{4}-2x+k2π\\x-\dfrac{π}{4}=π-(\dfrac{π}{4}-2x)+k2π\end{array}\right. \ (k\in Z)$
$⇔\left[\begin{array}{l}3x=\dfrac{π}{2}+k2π\\-x=π+k2π\end{array}\right.\ (k\in Z)$
$⇔\left[\begin{array}{l}x=\dfrac{π}{6}+\dfrac{k2π}{3}\\x=-π-k2π\end{array}\right.\ (k\in Z)$