Phương trình cos2(x+ $\frac{pi}{3}$ ) + 4 cos ( $\frac{pi}{6}$ - x) = $\frac{5}{2}$ có bao nhiêu nghiệm thuộc (-2019pi; 2019pi)
1 câu trả lời
Đáp án:
4038
Giải thích các bước giải:
\(\eqalign{ & \cos 2\left( {x + {\pi \over 3}} \right) + 4\cos \left( {{\pi \over 6} - x} \right) = {5 \over 2} \cr & \Leftrightarrow \cos 2\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) = {5 \over 2} \cr & \Leftrightarrow 1 - 2{\sin ^2}\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) = {5 \over 2} \cr & \Leftrightarrow - 2{\sin ^2}\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) - {3 \over 2} = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin \left( {x + {\pi \over 3}} \right) = {3 \over 2}\,\,\left( {loai} \right) \hfill \cr \sin \left( {x + {\pi \over 3}} \right) = {1 \over 2}\,\,\,\left( {tm} \right) \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x + {\pi \over 3} = {\pi \over 6} + k2\pi \hfill \cr x + {\pi \over 3} = {{5\pi } \over 6} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = - {\pi \over 6} + k2\pi \hfill \cr x = {\pi \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr & Xet\,\,ho\,\,nghiem\,\,x = - {\pi \over 6} + k2\pi \cr & - 2019\pi < x < 2019\pi \cr & \Rightarrow - 2019\pi < - {\pi \over 6} + k2\pi < 2019\pi \cr & \Leftrightarrow - 1009,41 < k < 1009,58 \cr & k \in Z \Rightarrow k \in \left\{ { - 1009; - 1008;...;1009} \right\} \Rightarrow Co\,\,2019\,\,nghiem \cr & Xet\,\,ho\,\,nghiem\,\,x = {\pi \over 2} + k2\pi \cr & - 2019\pi < x < 2019\pi \cr & \Rightarrow - 2019\pi < {\pi \over 2} + k2\pi < 2019\pi \cr & \Leftrightarrow - 1009,75 < k < 1009,25 \cr & k \in Z \Rightarrow k \in \left\{ { - 1009; - 1008;...;1009} \right\} \Rightarrow Co\,\,2019\,\,nghiem \cr & Vay\,\,co\,\,tat\,\,ca\,\,2019.2 = 4038\,\,nghiem\,\,thoa\,\,man. \cr} \)