Đáp án:
\(\eqalign{
& a)\,\,x = \pm {{2\pi } \over 3} + k2\pi \,\,\left( {k \in Z} \right) \cr
& b)\,\,x = - {\alpha \over 2} + {\pi \over 4} + k\pi \,\,\left( {k \in Z} \right)\,\,\,voi\,\,\,\left\{ \matrix{
\cos \alpha = {{12} \over {13}} \hfill \cr
\sin \alpha = {5 \over {13}} \hfill \cr} \right. \cr} \)
Giải thích các bước giải:
\(\eqalign{
& a)\,\,\cos 2x + 9\cos x + 5 = 0 \cr
& \Leftrightarrow 2{\cos ^2}x - 1 + 9\cos x + 5 = 0 \cr
& \Leftrightarrow 2{\cos ^2}x + 9\cos x + 4 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos x = - {1 \over 2} \hfill \cr
\cos x = - 4\,\,\,\left( {loai} \right) \hfill \cr} \right. \cr
& \Leftrightarrow x = \pm {{2\pi } \over 3} + k2\pi \,\,\left( {k \in Z} \right) \cr
& b)\,\,5\cos 2x + 12\sin 2x = 13 \cr
& \Leftrightarrow {5 \over {13}}\cos 2x + {{12} \over {13}}\sin 2x = 1 \cr
& Dat\,\,\left\{ \matrix{
\cos \alpha = {{12} \over {13}} \hfill \cr
\sin \alpha = {5 \over {13}} \hfill \cr} \right. \cr
& \Rightarrow \sin 2x\cos \alpha + \cos 2x\sin \alpha = 1 \cr
& \Leftrightarrow \sin \left( {2x + \alpha } \right) = 1 \cr
& \Leftrightarrow 2x + \alpha = {\pi \over 2} + k2\pi \cr
& \Leftrightarrow 2x = - \alpha + {\pi \over 2} + k2\pi \cr
& \Leftrightarrow x = - {\alpha \over 2} + {\pi \over 4} + k\pi \,\,\left( {k \in Z} \right) \cr} \)
