Đáp án:
Phương trình có tập nghiệm là:
$S=\Bigg\{-\dfrac{7\pi}{8};-\dfrac{3\pi}{8};\dfrac{\pi}{8};\dfrac{5\pi}{8};-\dfrac{3\pi}{4};\dfrac{\pi}{4}\Bigg\}$
Giải thích các bước giải:
$sin3x-cosx=0$
$↔ sin3x=cosx$
$↔ sin3x=sin\Bigg(\dfrac{\pi}{2}-x\Bigg)$
$↔ \left[ \begin{array}{l}3x=\dfrac{\pi}{2}-x+k2\pi\\3x=x+\dfrac{\pi}{2}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{\pi}{4}+k\pi\end{array} \right.$ ($k∈\mathbb{Z}$)
Vì $x∈(-\pi;\pi)$ nên:
+) $-\pi<\dfrac{\pi}{8}+k\dfrac{\pi}{2}<\pi$
$↔ -\dfrac{9}{4}<k<\dfrac{7}{4}$
Vì $k∈\mathbb{Z}$ nên $k∈\{-2;-1;0;1\}$
$→ x∈\Bigg(-\dfrac{7\pi}{8};-\dfrac{3\pi}{8};\dfrac{\pi}{8};\dfrac{5\pi}{8}\Bigg)$
+) $-\pi<\dfrac{\pi}{4}+k\pi<\pi$
$↔ -\dfrac{5}{4}<k<\dfrac{3}{4}$
Vì $k∈\mathbb{Z}$ nên $k∈\{-1;0\}$
$→ x∈\Bigg(-\dfrac{3\pi}{4};\dfrac{\pi}{4}\Bigg)$
