Sinx +√3 cosx=2 Cos 7x-√3sin7x=-√2 2cos 3x+√3 sinx+cosx=0

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1. Đáp án:

   \(\eqalign{
   & a)\,\,x = {\pi \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr
   & b)\,\,\left[ \matrix{
   x = {{5\pi } \over {84}} + {{k2\pi } \over 7} \hfill \cr
   x = – {{13\pi } \over {84}} + {{k2\pi } \over 7} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr
   & c)\,\,\left[ \matrix{
   x = {\pi \over 3} + k\pi \hfill \cr
   x = {\pi \over 3} + {{k\pi } \over 2} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)

   Giải thích các bước giải:

   $$\eqalign{
   & a)\,\,\sin x + \sqrt 3 \cos x = 2 \cr
   & \Leftrightarrow {1 \over 2}\sin x + {{\sqrt 3 } \over 2}\cos x = 1 \cr
   & \Leftrightarrow \sin x\cos {\pi \over 3} + \cos x\sin {\pi \over 3} = 1 \cr
   & \Leftrightarrow \sin \left( {x + {\pi \over 3}} \right) = 1 \cr
   & \Leftrightarrow x + {\pi \over 3} = {\pi \over 2} + k2\pi \cr
   & \Leftrightarrow x = {\pi \over 6} + k2\pi \,\,\left( {k \in Z} \right) \cr
   & b)\,\,\cos 7x – \sqrt 3 \sin 7x = – \sqrt 2 \cr
   & \Leftrightarrow {1 \over 2}\cos 7x – {{\sqrt 3 } \over 2}\sin 7x = – {{\sqrt 2 } \over 2} \cr
   & \Leftrightarrow \cos 7x\cos {\pi \over 3} – \sin 7x\sin {\pi \over 3} = – {{\sqrt 2 } \over 2} \cr
   & \Leftrightarrow \cos \left( {7x + {\pi \over 3}} \right) = – {{\sqrt 2 } \over 2} \cr
   & \Leftrightarrow \left[ \matrix{
   7x + {\pi \over 3} = {{3\pi } \over 4} + k2\pi \hfill \cr
   7x + {\pi \over 3} = – {{3\pi } \over 4} + k2\pi \hfill \cr} \right. \cr
   & \Leftrightarrow \left[ \matrix{
   7x = {{5\pi } \over {12}} + k2\pi \hfill \cr
   7x = – {{13\pi } \over {12}} + k2\pi \hfill \cr} \right. \cr
   & \Leftrightarrow \left[ \matrix{
   x = {{5\pi } \over {84}} + {{k2\pi } \over 7} \hfill \cr
   x = – {{13\pi } \over {84}} + {{k2\pi } \over 7} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr
   & c)\,\,2\cos 3x + \sqrt 3 \sin x + \cos x = 0 \cr
   & \Leftrightarrow {{\sqrt 3 } \over 2}\sin x + {1 \over 2}\cos x = – \cos 3x \cr
   & \Leftrightarrow \sin x\cos {\pi \over 6} + \cos x\sin {\pi \over 6} = – \cos 3x \cr
   & \Leftrightarrow \sin \left( {x + {\pi \over 6}} \right) = \sin \left( {3x – {\pi \over 2}} \right) \cr
   & \Leftrightarrow \left[ \matrix{
   x + {\pi \over 6} = 3x – {\pi \over 2} + k2\pi \hfill \cr
   x + {\pi \over 6} = \pi – 3x + {\pi \over 2} + k2\pi \hfill \cr} \right. \cr
   & \Leftrightarrow \left[ \matrix{
   2x = {{2\pi } \over 3} + k2\pi \hfill \cr
   4x = {{4\pi } \over 3} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
   x = {\pi \over 3} + k\pi \hfill \cr
   x = {\pi \over 3} + {{k\pi } \over 2} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$

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