a)

PT $\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x+2(\sin ^2x+\cos ^2x)=2$

$\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x=0$

$\Leftrightarrow (\sin x-3\cos x)(\sin x-\cos x)=0$

Nếu $\sin x-3\cos x=0$. Dễ thấy $\sin x, \cos x\neq 0$ nên $\tan x=\frac{\sin x}{\cos x}=3$

$\Rightarrow x=k\pi +\tan ^{-1}(3)$ với $k$ nguyên

Nếu $\sin x=\cos x$ thì tương tự ta có $\tan x=1\Rightarrow x=\pi (k+\frac{1}{4})$ với $k$ nguyên

b)
PT $\Leftrightarrow 25(\sin ^2x+\cos ^2x)+30\sin x\cos x-16\cos ^2x=25$

$\Leftrightarrow 30\sin x\cos x-16\cos ^2x=0$

$\Leftrightarrow \cos x(15\sin x-8\cos x)=0$

Nếu $\cos x=0\Rightarrow x=\pi (k+\frac{1}{2})$ với $k$ nguyên

Nếu $15\sin x-8\cos x=0$

Dễ thấy $\cos x\neq 0$ nên suy ra $\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}$

$\Rightarrow x=k\pi +\tan ^{-1}(\frac{8}{15})$ với $k$ nguyên

c) \(\left\{\begin{matrix} \sin x+\cos x=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (\sin x+\cos x)^2=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\)

\(\Rightarrow 2\sin x\cos x=0\Leftrightarrow \sin 2x=0\Rightarrow x=\frac{k}{2}\pi\) với $k$ nguyên.

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}-\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}+\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{3x}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{2}{x^2}}=0\)

\(-1\le\sin2x\le1\Rightarrow\dfrac{-5}{x^2+2}\le\dfrac{5\sin2x}{x^2+2}\le\dfrac{5}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}-\dfrac{5}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{5}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}=0\)

\(0\le\cos^2x\le1\Rightarrow0\le\dfrac{\cos^2x}{x^2+2}\le\dfrac{1}{x^2+2}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}=0\)

\(\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{3x-5\sin2x+\cos^2x}{x^2+2}=0\)

cos 2 x   +   2 sin x . cos x   +   5 sin 2 x   =   2

Rõ ràng cosx = 0 không thỏa mãn phương trình. Với cosx ≠ 0, chia hai vế cho cos2x ta được:

1   +   2 tan x   +   5 tan 2 x   =   2 ( 1   +   tan 2 x ) ⇔   3 tan 2 x   +   2 tan x   -   1   =   0

ĐKXĐ: \(sinx\ne\pm1\)

\(\dfrac{3cos2x-2sinx+5}{2\left(1-sin^2x\right)}=0\)

\(\Leftrightarrow3\left(1-2sin^2x\right)-2sinx+5=0\)

\(\Leftrightarrow-6sin^2x-2sinx+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(loại\right)\\sinx=-\dfrac{4}{3}< -1\left(loại\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

5 sin 2 x   +   3 cos x   +   3   =   0     ⇔   5 ( 1   -   cos 2 x )   +   3 cos x   +   3   =   0     ⇔   5 cos 2 x   -   3 cos x   -   8   =   0     ⇔   ( cos x   +   1 ) ( 5 cos x   -   8 )   =   0     ⇔   cos x   =   - 1     ⇔   x   =   ( 2 k   +   1 ) π ,   k   ∈   Z

Đề là: \(2sin^22x-3cos2x+6sin^2x-9=0\) đúng không nhỉ?

\(\Leftrightarrow2\left(1-cos^22x\right)-3cos2x+3\left(1-cos2x\right)-9=0\)

\(\Leftrightarrow-2cos^22x-6cos2x-4=0\)

\(\Leftrightarrow cos^22x+3cos2x+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=-2\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow...\)