\(a)\sin \left( {x - {\pi \over 3}} \right) = {{\sqrt 3 } \over 2}\)
\(\Leftrightarrow \,\sin \left( {x - {\pi \over 3}}\right) = \sin {\pi \over 3}\)
\(\Leftrightarrow \left[ {\matrix{{x - {\pi \over 3} = {\pi \over 3} + k2\pi } \cr {x - {\pi \over 3} = \pi - {\pi \over 3} + k2\pi } \cr} } \right. \)
\(\Leftrightarrow \left[ {\matrix{{x = {{2\pi } \over 3} + k2\pi } \cr {x = \pi + k2\pi } \cr} (k \in Z)} \right.\)
Vậy phương trình có nghiệm là: \(x = \pi + k2\pi ,x = \dfrac{{2\pi }}{3} + k2\pi (k \in \mathbb{Z})\)
\(b)\cos \left( {2x + {{30}^0}} \right) = {1 \over 2}\)
\(\Leftrightarrow \,\cos (2x + {30^0}) = \cos {60^0}\)
\(\Leftrightarrow \left[ {\matrix{{2x + {{30}^0} = {{60}^0} + k{{360}^0}} \cr {2x + {{30}^0} = - {{60}^0} + k{{360}^0}} \cr} } \right. \)
\(\Leftrightarrow \left[ {\matrix{{x = {{15}^0} + k{{360}^0}} \cr {x = - {{45}^0} + k{{360}^0}} \cr} (k} \right. \in Z)\)
Vậy phương trình có nghiệm là: \(x = {15^0} + k{360^0}\), \(x = - {45^0} + k{360^0}(k \in \mathbb{Z})\)
\(\begin{array}{l}c)\,{\cos ^2}x - 3\sin x = 1\\ \Leftrightarrow 1 - {\sin ^2}x - 3\sin x = 1\\ \Leftrightarrow {\sin ^2}x + 3\sin x = 0\\ \Leftrightarrow \sin x(\sin x + 3) = 0\\\end{array}\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = 0}\\{\sin x = - 3}\end{array}} \right. \Leftrightarrow \sin x = 0\)
(\(\sin x = - 3\) vô nghiệm vì \( - 1 \le \sin x \le 1\) )
\( \Leftrightarrow x = k\pi (k \in \mathbb{Z})\)
Vậy phương trình có nghiệm là: \(x = k\pi (k \in \mathbb{Z})\)
\(d)\,\,\sin 3x + 4\cos 2x - \sin x = 0\)
\(\Leftrightarrow 3\sin x - 4{\sin ^3}x + 4(1 - 2{\sin ^2}x) - \sin x = 0\)
\(\Leftrightarrow 2{\sin ^3}x + 4{\sin ^2}x - \sin x - 2 = 0\,(1)\)
Đặt \(\sin x = t\,\,(\left| t \right| \le 1)\)
Khi đó phương trình (1) trở thành \(2{t^3} + 4{t^2} - t - 2 = 0\)
\(\Leftrightarrow (t + 2)(2{t^2} - 1) = 0\)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = - 2\,(KTM)}\\{t = \pm \dfrac{1}{{\sqrt 2 }}\,(TM)}\end{array}} \right.\)
Với \(t = \dfrac{1}{{\sqrt 2 }} \Rightarrow \sin x = \dfrac{1}{{\sqrt 2 }}\)
\(\Leftrightarrow \sin x = \sin \dfrac{\pi }{4}\)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + k2\pi }\\{x = \dfrac{{3\pi }}{4} + k2\pi }\end{array}} \right.\,(k \in \mathbb{Z})\)
Với \(t = - \dfrac{1}{{\sqrt 2 }} \Rightarrow \sin x = - \dfrac{1}{{\sqrt 2 }}\)
\(\Leftrightarrow \sin x = \sin ( - \dfrac{\pi }{4})\)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \dfrac{\pi }{4} + k2\pi }\\{x = \dfrac{{5\pi }}{4} + k2\pi }\end{array}} \right.\,(k \in \mathbb{Z})\)
Vậy phương trình có nghiệm là: \(x = \pm \dfrac{\pi }{4} + k2\pi \); \(x = \dfrac{{3\pi }}{4} + k2\pi \); \(x = \dfrac{{5\pi }}{4} + k2\pi (k \in \mathbb{Z})\)
Bài 2:
\(\begin{array}{l}y = 6\sin 2x - 8\cos 2x - 2\\\,\,\,\, = 10(\dfrac{3}{5}\sin 2x - \dfrac{4}{5}\cos 2x) - 2\end{array}\)
Đặt \(\cos \alpha = \dfrac{3}{5};\,\,\,\,\,\,\sin \alpha = \dfrac{4}{5}\)
Khi đó
\(\begin{array}{l}y = 10(\cos \alpha \sin 2x - \sin \alpha \cos 2x) - 2\\\,\,\,\, = 10\sin (2x - \alpha ) - 2\end{array}\)
Ta có: \( - 1 \le \sin (2x - \alpha ) \le 1\)
\(\Leftrightarrow - 10 \le 10\sin (2x - \alpha ) \le 10\)
\(\Leftrightarrow - 12 \le y \le 8\,(\forall x \in \mathbb{R})\)
Vậy \(\min y = - 12\) khi \(\sin (2x - \alpha ) = - 1\)
\(\Leftrightarrow 2x - \alpha = \dfrac{{ - \pi }}{2} + k2\pi \)
\(\Leftrightarrow x = \dfrac{{ - \pi }}{4} + \dfrac{\alpha }{2} + k\pi (k \in \mathbb{Z})\)
\(\max y = 8\,\,\,\) khi \(\,\sin (2x - \alpha ) = 1 \)
\(\Leftrightarrow 2x - \alpha = \dfrac{\pi }{2} + k2\pi \)
\(\Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{\alpha }{2} + k\pi (k \in \mathbb{Z})\)
