Đáp án:
$S=\{\ \dfrac{\pi}{6}+k2\pi ; \dfrac{7\pi}{12}+k2\pi |k\in Z \}$
Giải thích các bước giải:
$sinx+cosx+\sqrt{3}(sinx-cosx)=2$
$sinx+\sqrt{3}sinx+cosx-\sqrt{3}cosx=2$
$(1+\sqrt{3})sinx+(1-\sqrt{3})cosx=2$
Chia hai vế cho $2\sqrt{2}$ ta có :
$\dfrac{1+\sqrt{3}}{2\sqrt{2}}sinx+\dfrac{1-\sqrt{3}}{2\sqrt{2}}cosx=\dfrac{1}{\sqrt{2}}$
$cos\dfrac{\pi}{12}sinx+sin\dfrac{\pi}{12}cosx=\dfrac{1}{\sqrt{2}}$
$sin(\dfrac{\pi}{12}+x)=\dfrac{1}{\sqrt{2}}$
\(\left[ \begin{array}{l}\dfrac{\pi}{12}+x=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{12}+x=\pi-\dfrac{\pi}{4}+k2\pi\end{array} \right.\)
\(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{array} \right.,k\in Z\)
Vậy $S=\{\ \dfrac{\pi}{6}+k2\pi ; \dfrac{7\pi}{12}+k2\pi |k\in Z \}$
