Đáp án:
Vay PT co 2 nghiem \(\left[ \begin{array}{l}x=\frac{k2π}{3}\\x=\frac{π}{3}+ k2π \end{array} \right.\)
Giải thích các bước giải:
3cosx-sin2x=√3cos2x+√3sinx
⇔ 3cosx-√3sinx=√3cos2x+sin2x
⇔ $\frac{√3}{2}$ cosx-$\frac{1}{2}$ sinx= $\frac{√3}{2}$ cos2x-$\frac{1}{2}$sin2x
⇔sin$\frac{π}{3}$cosx-cos $\frac{π}{3}$sinx =sin$\frac{π}{3}$cos2x-cos $\frac{π}{3}$sin2x
⇔sin($\frac{π}{3}$-x)=sin($\frac{π}{3}$+2x)
⇔ \(\left[ \begin{array}{l}π/3-x=\frac{π}{3}+2x+k2π\\π/3-x=π-\frac{π}{3}-2x+k2π\end{array} \right.\)
⇔
\(\left[ \begin{array}{l}x=\frac{k2π}{3}\\x=\frac{π}{3}+ k2π \end{array} \right.\)