Đáp án:

c) \(S = \left\{ {\dfrac{\pi }{4} + k\pi ,\,\,\arctan 3 + k\pi ,\,\,k \in Z} \right\}\).

d) \(S = \left\{ {k2\pi ,\,\, \pm \dfrac{\pi }{3} + k2\pi ,\,\,k \in Z} \right\}\).

e) \(S = \left\{ { - \dfrac{{5\pi }}{6} + k2\pi ,\,\,k \in Z} \right\}\).

Giải thích các bước giải:

c) \(3{\sin ^2}x - 4\sin x\cos x + 5{\cos ^2}x = 2\).

TH1: \(\cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\left( {k \in Z} \right)\).

Khi đó \({\sin ^2}x = 1\).

\( \Rightarrow 1 - 4.0 + 5.0 = 2\) (Vô nghiệm).

TH2: \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\left( {k \in Z} \right)\).

Chia cả 2 vế cho \({\cos ^2}x\) ta được:

\(\begin{array}{l}3{\tan ^2}x - 4\tan x + 5 = 2\left( {1 + {{\tan }^2}x} \right)\\ \Leftrightarrow {\tan ^2}x - 4\tan x + 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x = \arctan 3 + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\)

Vậy \(S = \left\{ {\dfrac{\pi }{4} + k\pi ,\,\,\arctan 3 + k\pi ,\,\,k \in Z} \right\}\).

d) \(2{\cos ^2}x - 3\cos x + 1 = 0\)

\(\begin{array}{l} \Leftrightarrow 2{\cos ^2}x - 2\cos x - \cos x + 1 = 0\\ \Leftrightarrow 2\cos x\left( {\cos x - 1} \right) - \left( {\cos x - 1} \right) = 0\\ \Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 1\\\cos x = \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x =  \pm \dfrac{\pi }{3} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\)

Vậy \(S = \left\{ {k2\pi ,\,\, \pm \dfrac{\pi }{3} + k2\pi ,\,\,k \in Z} \right\}\).

e) \(\sqrt 3 \cos x + \sin x = 2\)

Chia cả 2 vế cho \(\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}  = 2\) ta được:

\(\begin{array}{l}\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x =  - 1\\ \Leftrightarrow \sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3} =  - 1\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) =  - 1\\ \Leftrightarrow x + \dfrac{\pi }{3} =  - \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x =  - \dfrac{{5\pi }}{6} + k2\pi \,\,\left( {k \in Z} \right)\end{array}\)

Vậy \(S = \left\{ { - \dfrac{{5\pi }}{6} + k2\pi ,\,\,k \in Z} \right\}\).