Câu 1:
\(a)\,2{\sin ^2}x + 5\cos x + 1 = 0 \)
\(\Leftrightarrow 2(1 - {\cos ^2}x) + 5\cos x + 1 = 0 \)
\(\Leftrightarrow 2{\cos ^2}x - 5\cos x - 3 = 0\)
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: \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \)
\((1)\,\,\,\,\,\, \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \sqrt 3 }\\{\tan x = - 1}\end{array}} \right.\)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \tan \dfrac{\pi }{3}}\\{\tan x = \tan (\dfrac{{ - \pi }}{4})}\end{array}} \right. \)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{3} + k\pi \,\,(TM)}\\{x = \dfrac{{ - \pi }}{4} + k\pi \,\,(KTM)}\end{array}} \right.\,\,(k \in \mathbb{Z})\)
\(c) \; {\sin ^2}x = \dfrac{1}{2} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = \dfrac{1}{{\sqrt 2 }}\,\,\,\,\,(1)}\\{\sin x = - \dfrac{1}{{\sqrt 2 }}\,\,\,(2)}\end{array}} \right.\)
\(\begin{array}{l}(1) \Leftrightarrow \sin x = \sin \dfrac{\pi }{4}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + k2\pi }\\{x = \dfrac{{3\pi }}{4} + k2\pi }\end{array}\,\,(} \right.k \in \mathbb{Z})\\(2) \Leftrightarrow \sin x = \sin ( - \dfrac{\pi }{4})\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{ - \pi }}{4} + k2\pi }\\{x = \dfrac{{5\pi }}{4} + k2\pi }\end{array}} \right.\,\,(k \in \mathbb{Z})\end{array}\)
\(d) \,\cos 2x - 3\cos x = 4{\cos ^2}\dfrac{x}{2}\)
\(\Leftrightarrow 2{\cos ^2}x - 1 - 3\cos x = 4\dfrac{{1 = \cos 2}}{2} \)
\(\Leftrightarrow 2{\cos ^2}x - 5\cos x - 3 = 0\)
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 \( \Leftrightarrow {m^2} + {(m + 2)^2} \ge {2^2} \)
\(\Leftrightarrow {m^2} + {m^2} + 4m + 4 \ge 4\)
\( \Leftrightarrow 2{m^2} + 4m \ge 0 \)
\(\Leftrightarrow m \in \left( { - \infty ; - 2} \right] \cup \left[ {0; + \infty } \right)\)
Vậy với \(m \in \left( { - \infty ; - 2} \right] \cup \left[ {0; + \infty } \right)\)thì phương trình có nghiệm.
