Đề bài
Câu 1: Giải các phương trình sau:
a] \[2{\sin ^2}x + 5\cos x + 1 = 0\]
b] \[{\tan ^2}x + \left[ {1 - \sqrt 3 } \right]\tan x - \sqrt 3 = 0\]
c] \[{\sin ^2}x = \dfrac{1}{2}\]
d] \[\cos 2x - 3\cos x = 4{\cos ^2}\dfrac{x}{2}\]
Câu 2:
a] Giải phương trình sau: \[\cos x + \sqrt 3 \sin x = \sqrt 2 \]
b] Tìm m để phương trình : \[m\cos x + [m + 2]\sin x = 2\] có nghiệm.
Lời giải chi tiết
Câu 1:
\[a]\,2{\sin ^2}x + 5\cos x + 1 = 0 \]
\[\Leftrightarrow 2[1 - {\cos ^2}x] + 5\cos x + 1 = 0 \]
\[\begin{array}{l}
\Leftrightarrow 2 - 2{\cos ^2}x + 5\cos x + 1 = 0\\
\Leftrightarrow - 2{\cos ^2}x + 5\cos x + 3 = 0
\end{array}\]
\[\Leftrightarrow 2{\cos ^2}x - 5\cos x - 3 = 0\]
Đặt: \[t = \cos x\,\,[ - 1 \le t \le 1]\]
Khi đó phương trình trở thành: \[2{t^2} - 5t - 3 = 0 \]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = 3\,\,[ktm]}\\{t = - \dfrac{1}{2}\,\,\,[tm]}\end{array}} \right.\]
Với \[t = - \dfrac{1}{2} \Rightarrow \cos x = - \dfrac{1}{2} \]
\[\Leftrightarrow \cos x = \cos \dfrac{{2\pi }}{3}\]
\[\Leftrightarrow x = \pm \dfrac{{2\pi }}{3} = k2\pi \]
\[b] \,\,{\tan ^2}x + \left[ {1 - \sqrt 3 } \right]\tan x - \sqrt 3 = 0\,\,\,\,\,\,\,\,[1]\]
ĐK: \[\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \]
\[[1]\,\,\,\,\,\, \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \sqrt 3 }\\{\tan x = - 1}\end{array}} \right.\]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \tan \dfrac{\pi }{3}}\\{\tan x = \tan [\dfrac{{ - \pi }}{4}]}\end{array}} \right. \]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{3} + k\pi \,\,[TM]}\\{x = \dfrac{{ - \pi }}{4} + k\pi \,\,[KTM]}\end{array}} \right.\,\,[k \in \mathbb{Z}]\]
\[c] \; {\sin ^2}x = \dfrac{1}{2} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = \dfrac{1}{{\sqrt 2 }}\,\,\,\,\,[1]}\\{\sin x = - \dfrac{1}{{\sqrt 2 }}\,\,\,[2]}\end{array}} \right.\]
\[\begin{array}{l}[1] \Leftrightarrow \sin x = \sin \dfrac{\pi }{4}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + k2\pi }\\{x = \dfrac{{3\pi }}{4} + k2\pi }\end{array}\,\,[} \right.k \in \mathbb{Z}]\\[2] \Leftrightarrow \sin x = \sin [ - \dfrac{\pi }{4}]\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{ - \pi }}{4} + k2\pi }\\{x = \dfrac{{5\pi }}{4} + k2\pi }\end{array}} \right.\,\,[k \in \mathbb{Z}]\end{array}\]
Cách khác:
\[\begin{array}{l}
{\sin ^2}x = \frac{1}{2} \Leftrightarrow \frac{{1 - \cos 2x}}{2} = \frac{1}{2}\\
\Leftrightarrow 1 - \cos 2x = 1 \Leftrightarrow \cos 2x = 0\\
\Leftrightarrow 2x = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}
\end{array}\]
\[d] \,\cos 2x - 3\cos x = 4{\cos ^2}\dfrac{x}{2}\]
\[\Leftrightarrow 2{\cos ^2}x - 1 - 3\cos x = 4.\dfrac{{1 + \cos x}}{2} \]
\[ \Leftrightarrow 2{\cos ^2}x - 3\cos x - 1 = 2 + 2\cos x\]
\[\Leftrightarrow 2{\cos ^2}x - 5\cos x - 3 = 0\]
Đặt: \[t = \cos x\,\,[ - 1 \le t \le 1]\] khi đó phương trình trở thành: \[2{t^2} - 5t - 3 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = 3\,\,[ktm]}\\{t = - \dfrac{1}{2}\,\,\,[tm]}\end{array}} \right.\]
Với \[t = - \dfrac{1}{2} \Rightarrow \cos x = - \dfrac{1}{2}\]
\[\Leftrightarrow \cos x = \cos \dfrac{{2\pi }}{3} \Leftrightarrow x = \pm \dfrac{{2\pi }}{3} = k2\pi \]
Câu 2:
\[\begin{array}{l}a] \cos x + \sqrt 3 \sin x = \sqrt 2 \\ \Leftrightarrow \dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x = \dfrac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin [\dfrac{\pi }{6} + x] = \sin \dfrac{\pi }{4}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \dfrac{\pi }{6} = \dfrac{\pi }{4} + k2\pi }\\{x + \dfrac{\pi }{6} = \pi - \dfrac{\pi }{4} + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{12}} + k2\pi }\\{x = \dfrac{{7\pi }}{{12}} + k2\pi }\end{array}} \right.\,\,[k \in \mathbb{Z}]\end{array}\]
\[b] \;m\cos x + [m + 2]\sin x = 2 [1]\]
Để PT[1] có nghiệm \[ \Leftrightarrow {m^2} + {[m + 2]^2} \ge {2^2} \]
\[\Leftrightarrow {m^2} + {m^2} + 4m + 4 \ge 4\]
\[ \Leftrightarrow 2{m^2} + 4m \ge 0 \]
\[ \Leftrightarrow \left[ \begin{array}{l}
m \ge 0\\
m \le - 2
\end{array} \right.\]
Vậy với \[m \in \left[ { - \infty ; - 2} \right] \cup \left[ {0; + \infty } \right]\] thì phương trình có nghiệm.