Câu a)
Đặt \(2x=a\). PT trở thành:
\(2\sin ^2a+\sin 3a-1=\sin a\)
\(\Leftrightarrow 2\sin ^2a+\sin (a+2a)-1-\sin a=0\)
\(\Leftrightarrow 2\sin ^2a+\sin a\cos 2a+\cos a\sin 2a-1-\sin a=0\)
\(\Leftrightarrow 2\sin ^2a+\sin a\cos 2a+2\cos ^2a\sin a-1-\sin a=0\)
\(\Leftrightarrow (2\sin ^2a-1)+\sin a\cos 2a+\sin a(2\cos ^2a-1)=0\)
\(\Leftrightarrow -\cos 2a+\sin a\cos 2a+\sin a\cos 2a=0\)
\(\Leftrightarrow \cos 2a(-1+2\sin a)=0\)
\(\Rightarrow \left[\begin{matrix} \cos 2a=0(1)\\ \sin a=\frac{1}{2}(2)\end{matrix}\right.\)
Từ (1) \(\Rightarrow 2a=\frac{\pi}{2}+k\pi (k\in\mathbb{Z})\)\(\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
Từ (2) \(\Rightarrow \left[\begin{matrix} a=\frac{\pi}{6}+2k\pi \rightarrow x=\frac{\pi}{12}+k\pi \\ a=\frac{5}{6}\pi+2k\pi \rightarrow x=\frac{5\pi}{12}+k\pi \end{matrix}\right.\)
Bài 2:
\(\sin 2x+\sin 6x+2\sin ^2x-1=0\)
\(\Leftrightarrow \sin 2x+\sin 6x-\cos 2x=0\)
\(\Leftrightarrow \sin 2x+\sin 4x\cos 2x+\cos 4x\sin 2x-\cos 2x=0\)
\(\Leftrightarrow \sin a+\sin 2a\cos a+\cos 2a\sin a-\cos a=0\)
\(\Leftrightarrow \sin a(1+\cos 2a)+\sin 2a\cos a-\cos a=0\)
\(\Leftrightarrow \sin a.2\cos ^2a+\sin 2a\cos a-\cos a=0\)
\(\Leftrightarrow \cos a(2\sin 2a-1)=0\)
\(\Rightarrow \left[\begin{matrix} \cos a=0(1)\\ \sin 2a=\frac{1}{2}(2)\end{matrix}\right.\)
Từ (1)\(\Rightarrow a=\frac{\pi}{2}+k\pi \Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
Từ (2) \(\Rightarrow \left[\begin{matrix} 2a=\frac{\pi}{6}+2k\pi \rightarrow x=\frac{\pi}{24}+\frac{k\pi}{2}\\ 2a=\frac{5\pi}{6}+2k\pi \rightarrow x=\frac{5\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)
