Giúp em bài này với ạ Mong được giải chi tiết ???????????? (2 sinx – 1)(2sin2x +1)=3 – 4 cos^x 01/10/2021 Bởi Maria Giúp em bài này với ạ Mong được giải chi tiết ???????????? (2 sinx – 1)(2sin2x +1)=3 – 4 cos^x
$$\eqalign{ & \left( {2\sin x – 1} \right)\left( {2\sin 2x + 1} \right) = 3 – 4{\cos ^2}x \cr & \Leftrightarrow 4\sin x\sin 2x + 2\sin x – 2\sin 2x – 1 = 3 – 4{\cos ^2}x \cr & \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 2\sin 2x – 4 + 4{\cos ^2}x = 0 \cr & \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 2\sin 2x – 4\left( {1 – {{\cos }^2}x} \right) = 0 \cr & \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 4\sin x\cos x – 4{\sin ^2}x = 0 \cr & \Leftrightarrow 2\sin x\left( {4\sin x\cos x + 1 – 2\cos x – 2\sin x} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin x = 0\,\,\left( 1 \right) \hfill \cr 4\sin x\cos x + 1 – 2\sin x – 2\cos x = 0\,\,\,\left( 2 \right) \hfill \cr} \right. \cr & \left( 1 \right) \Leftrightarrow x = k\pi \,\,\left( {k \in Z} \right) \cr & \left( 2 \right) \Leftrightarrow 4\sin x\cos x – 2\left( {\sin x + \cos x} \right) + 1 = 0 \cr & Dat\,\,t = \sin x + \cos x\,\,\left( { – \sqrt 2 \le t \le \sqrt 2 } \right) \cr & \Rightarrow {t^2} = 1 + 2\sin x\cos x \cr & \Leftrightarrow 2\sin x\cos x = {t^2} – 1 \cr & PT:\,\,2\left( {{t^2} – 1} \right) – 2t + 1 = 0 \cr & \Leftrightarrow 2{t^2} – 2t – 1 = 0 \Leftrightarrow \left[ \matrix{ t = {{1 + \sqrt 3 } \over 2}\, \hfill \cr t = {{1 – \sqrt 3 } \over 2} \hfill \cr} \right.\,\,\left( {tm} \right) \cr & t = {{1 \pm \sqrt 3 } \over 2}\, \Leftrightarrow \sqrt 2 \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over 2} \cr & \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over {2\sqrt 2 }} \cr & \Leftrightarrow \left[ \matrix{ x + {\pi \over 4} = \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr x + {\pi \over 4} = \pi – \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = – {\pi \over 4} + \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr x = {{3\pi } \over 4} – \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$ Bình luận
$$\eqalign{
& \left( {2\sin x – 1} \right)\left( {2\sin 2x + 1} \right) = 3 – 4{\cos ^2}x \cr
& \Leftrightarrow 4\sin x\sin 2x + 2\sin x – 2\sin 2x – 1 = 3 – 4{\cos ^2}x \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 2\sin 2x – 4 + 4{\cos ^2}x = 0 \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 2\sin 2x – 4\left( {1 – {{\cos }^2}x} \right) = 0 \cr
& \Leftrightarrow 8{\sin ^2}x\cos x + 2\sin x – 4\sin x\cos x – 4{\sin ^2}x = 0 \cr
& \Leftrightarrow 2\sin x\left( {4\sin x\cos x + 1 – 2\cos x – 2\sin x} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin x = 0\,\,\left( 1 \right) \hfill \cr
4\sin x\cos x + 1 – 2\sin x – 2\cos x = 0\,\,\,\left( 2 \right) \hfill \cr} \right. \cr
& \left( 1 \right) \Leftrightarrow x = k\pi \,\,\left( {k \in Z} \right) \cr
& \left( 2 \right) \Leftrightarrow 4\sin x\cos x – 2\left( {\sin x + \cos x} \right) + 1 = 0 \cr
& Dat\,\,t = \sin x + \cos x\,\,\left( { – \sqrt 2 \le t \le \sqrt 2 } \right) \cr
& \Rightarrow {t^2} = 1 + 2\sin x\cos x \cr
& \Leftrightarrow 2\sin x\cos x = {t^2} – 1 \cr
& PT:\,\,2\left( {{t^2} – 1} \right) – 2t + 1 = 0 \cr
& \Leftrightarrow 2{t^2} – 2t – 1 = 0 \Leftrightarrow \left[ \matrix{
t = {{1 + \sqrt 3 } \over 2}\, \hfill \cr
t = {{1 – \sqrt 3 } \over 2} \hfill \cr} \right.\,\,\left( {tm} \right) \cr
& t = {{1 \pm \sqrt 3 } \over 2}\, \Leftrightarrow \sqrt 2 \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over 2} \cr
& \Leftrightarrow \sin \left( {x + {\pi \over 4}} \right) = {{1 \pm \sqrt 3 } \over {2\sqrt 2 }} \cr
& \Leftrightarrow \left[ \matrix{
x + {\pi \over 4} = \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr
x + {\pi \over 4} = \pi – \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = – {\pi \over 4} + \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr
x = {{3\pi } \over 4} – \arcsin \left( {{{1 \pm \sqrt 3 } \over {2\sqrt 2 }}} \right) + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$