Đáp án:

 a)$x = \dfrac{\pi }{3} + k\pi ;x = k\pi \left( {k \in Z} \right)$

b)$x = \pi  + k2\pi ;x = \dfrac{{2\pi }}{3} + k2\pi ;x = \dfrac{{ - 2\pi }}{3} + k2\pi \left( {k \in Z} \right)$

c)$x =  \pm \dfrac{\pi }{4} + k\pi ;x = \dfrac{{7\pi }}{{12}} + k2\pi ;x = \dfrac{{ - \pi }}{{12}} + k2\pi \left( {k \in Z} \right)$

d)$x = k2\pi \left( {k \in Z} \right)$

Giải thích các bước giải:

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

Vậy các họ nghiệm của phương trình là: $x = \dfrac{\pi }{3} + k\pi ;x = k\pi \left( {k \in Z} \right)$

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

Vậy các họ nghiệm của phương trình là:$x = \pi  + k2\pi ;x = \dfrac{{2\pi }}{3} + k2\pi ;x = \dfrac{{ - 2\pi }}{3} + k2\pi \left( {k \in Z} \right)$

$\begin{array}{l}
c)\sin 3x + \cos 3x - \sin x + \cos x = \sqrt 2 \cos 2x\\
 \Leftrightarrow 3\sin x - 4{\sin ^3}x + 4{\cos ^3}x - 3\cos x - \sin x + \cos x = \sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\
 \Leftrightarrow 2\left( {\sin x - \cos x} \right) - 4\left( {{{\sin }^3}x - {{\cos }^3}x} \right) + \sqrt 2 \left( {{{\sin }^2}x - {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {2 - 4\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right) + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {2 - 4\left( {1 + \sin x\cos x} \right) + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( { - 2 - 4\sin x\cos x + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( { - 2{{\left( {\sin x + \cos x} \right)}^2} + \sqrt 2 \left( {\sin x + \cos x} \right)} \right) = 0\\
 \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)\left( { - 2\left( {\sin x + \cos x} \right) + \sqrt 2 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
\sin x + \cos x = 0\\
\sin x + \cos x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x =  - 1\\
\cos \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{4} + k\pi \\
x - \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\
x - \dfrac{\pi }{4} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{ - \pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy các họ nghiệm của phương trình là:$x =  \pm \dfrac{\pi }{4} + k\pi ;x = \dfrac{{7\pi }}{{12}} + k2\pi ;x = \dfrac{{ - \pi }}{{12}} + k2\pi \left( {k \in Z} \right)$

$\begin{array}{l}
d)9 - 13\cos x + \dfrac{4}{{1 + {{\tan }^2}x}} = 0\left( {DK:x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)} \right)\\
 \Leftrightarrow 9 - 13\cos x + 4{\cos ^2}x = 0\\
 \Leftrightarrow \left( {\cos x - 1} \right)\left( {4\cos x - 9} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{9}{4}\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x = k2\pi \left( {k \in Z} \right)\left( {tm} \right)
\end{array}$

Vậy các họ nghiệm của phương trình là:$x = k2\pi \left( {k \in Z} \right)$

a, Ta có ` sqrt3.sin2x+sin(π/2 +2x) =1`

`<=> cos(2 x) + sqrt(3) sin(2 x) = 2 [1/2 cos(2 x) + 1/2 sqrt(3) sin(2 x)]`

`= 2 [sin(π/6) cos(2 x) + cos(π/6) sin(2 x)]`

`= 2 sin(2 x + π/6): 2 sin(2 x + π/6) = 1`

`<=> sin(2 x + π/6) = 1/2`

`<=>` \(\left[ \begin{array}{l}2 x + \frac{\pi}{6} = 2 π n + \frac{5\pi}{6} (n \in \mathbb{Z})\\2 x + \frac{\pi}{6} = 2 π n + \frac{\pi}{6}(n \in \mathbb{Z})\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}2 x  = 2 π n + \frac{2\pi}{3}(n \in \mathbb{Z})\\2 x = 2 π n (n \in \mathbb{Z})\end{array} \right.\)  

`<=>` \(\left[ \begin{array}{l}x  =  π n + \frac{\pi}{3}(n \in \mathbb{Z})\\ x =  π n (n \in \mathbb{Z}) \end{array} \right.\)  

Vậy phương trình có tập nghiệm

`S={π n +\frac{\pi}{3}   ;   π n | n \in \mathbb{Z}}`

b) Ta có: `cos2x+3cosx+2=0`

`<=> 1 + 3 cos(x) + 2 cos^2(x) = 0`

`<=> [cos(x) + 1] [2 cos(x) + 1] = 0`

`<=>` \(\left[ \begin{array}{l}cos(x) + 1 = 0\\ 2 cos(x) + 1 = 0 \end{array} \right.\)  

`<=>` \(\left[ \begin{array}{l}cos(x)  = -1\\ 2 cos(x) = -1 \end{array} \right.\)  

`<=>` \(\left[ \begin{array}{l}x = 2 π n + π (n \in \mathbb{Z})\\ cos(x) = \frac{-1}{2} \end{array} \right.\)  

`<=>` \(\left[ \begin{array}{l}x = 2 π n + π (n \in \mathbb{Z})\\ x = 2 π n + \frac{2\pi}{3}(n \in \mathbb{Z})\\x = 2 π n + \frac{4\pi}{3}(n \in \mathbb{Z})\end{array} \right.\)  

Vậy phương trình có tập nghiệm:

`S={2 π n + π   ;   2 π n + \frac{2\pi}{3} ; x = 2 π n + \frac{4\pi}{3} | n \in \mathbb{Z}}`

 c, Ta có: `sin3x+cos3x-sinx+cosx=\sqrt2 cos2x`

`<=> cos(x) - sqrt(2) cos(2 x) + cos(3 x) - sin(x) + sin(3 x) = 0`

`<=> 2 sin(-x + π/4) sin(x + π/4) [2 sqrt(2) sin(x + π/4) - sqrt(2)] = 0`

`<=> sin(-x + π/4) sin(x + π/4) [2 sqrt(2) sin(x + π/4) - sqrt(2)] = 0`

`<=>` \(\left[ \begin{array}{l}sin(-x + \frac{\pi}{4}) = 0\\ sin(x + \frac{\pi}{4}) = 0\\2 \sqrt2 sin(x + \frac{\pi}{4}) -\sqrt2 = 0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}-x + \frac{\pi}{4} = π n(n\in \mathbb{Z})\\ x + \frac{\pi}{4} = π n (n\in \mathbb{Z})\\\sqrt2 [2 sin(x + \frac{\pi}{4}) - 1] = 0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} - π n(n\in \mathbb{Z})\\ x   = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\2 sin(x + \frac{\pi}{4}) - 1 = 0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} - π n(n\in \mathbb{Z})\\ x   = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ sin(x + \frac{\pi}{4})  = \frac{1}{2}\end{array} \right.\)  
`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} - π n(n\in \mathbb{Z})\\ x   = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ x + \frac{\pi}{4}= 2 π n + \frac{5\pi}{6}(n\in \mathbb{Z})\\x + \frac{\pi}{4}= 2 π n + \frac{\pi}{6}(n\in \mathbb{Z})\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \frac{\pi}{4} - π n(n\in \mathbb{Z})\\ x   = π n-\frac{\pi}{4} (n\in \mathbb{Z})\\ x = 2 π n + \frac{7\pi}{12}(n\in \mathbb{Z})\\x = 2 π n - \frac{\pi}{12}(n\in \mathbb{Z})\end{array} \right.\) 

Vậy phương trình có tập nghiệm:

`S={\frac{\pi}{4} - π n   ;  π n-\frac{\pi}{4} ; x = 2 π n + \frac{7\pi}{12} ; 2 π n - \frac{\pi}{12} | n \in \mathbb{Z}}`

d) Đề bài là `(9-13cosx+4)/(1+tan^2x)=0` hay là như thế nào hả bạn? Sau mình bổ sung nhé.