Đáp án:

$S=\left \{ k\pi,\dfrac{\pi}{2}+k\pi \right \}$

Giải thích các bước giải:

 $\cos3x+\sin3x=2\sin2x+\sin x+\cos x\\
\Leftrightarrow (\cos x+\sin x)(\sin^2x+\sin x\cos x+\cos^2x)\\=2\sin2x+\sin x+\cos x\\
\Leftrightarrow (\cos x+\sin x)(1+\sin x\cos x)=2\sin2x+(\sin x+\cos x)\\
\Leftrightarrow (\cos x+\sin x)(1+\sin x\cos x)-2\sin2x-(\sin x+\cos x)=0\\
\Leftrightarrow (\sin x+\cos x)(1+\sin x\cos x-1)-4\sin x\cos x=0\\
\Leftrightarrow (\sin x+\cos x).\sin x\cos x-4\sin x\cos x=0\\
\Leftrightarrow \sin x\cos x(\sin x+\cos x-4)=0\\
\Leftrightarrow {\left[\begin{aligned}\sin x=0\\ \cos x=0\\ \sin x+\cos x-4=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=k\pi\\ x=\dfrac{\pi}{2}+k\pi\\ \sin x+\cos x=4\end{aligned}\right.}\\
+) \sin x+\cos x=4\\
\Leftrightarrow \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x=2\sqrt{2}\\
 \Leftrightarrow \cos\dfrac{\pi}{4}\sin x+\sin\dfrac{\pi}{4}\cos x=2\sqrt{2}\\
 \Leftrightarrow \cos\left ( x-\dfrac{\pi}{4} \right )=2\sqrt{2}(VN)$
Vì $2\sqrt{2}\notin [-1;1]$ nên phương trình vô nghiệm
Vậy $S=\left \{ k\pi,\dfrac{\pi}{2}+k\pi \right \}$

Đáp án:

$x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}\cos^3x + \sin^3x = 2\sin2x + \sin x + \cos x\\ \Leftrightarrow (\cos x + \sin x)(1 - \sin x\cos x) =4\sin x\cos x + \sin x + \cos x\\ Đặt\,\,t = \sin x + \cos x \qquad (|t| \leq \sqrt2)\\ \Rightarrow t^2 = 1 + 2\sin x\cos x\\ \Rightarrow t^2 - 1 = 2\sin x\cos x\\ \Rightarrow \dfrac{t^2 - 1}{2} = \sin x\cos x\\ \text{Phương trình trở thành:}\\ t.\left(1 - \dfrac{t^2 - 1}{2}\right)= 2(t^2 - 1) + t\\ \Leftrightarrow t^3 + 4t^2 - t - 4 = 0\\ \Leftrightarrow (t-1)(t+1)(t + 4) = 0\\\Leftrightarrow \left[\begin{array}{l}t = 1\qquad (nhận)\\t=-1\quad (nhận)\\t = -4\quad (loại)\end{array}\right.\\ \Leftrightarrow t = 1\\+) \quad Với\,\,t=1\,\,ta\,\,được:\\ \sin x + \cos x =1\\ \Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right) = 1\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\qquad (k \in \Bbb Z)\\+) \quad Với\,\,t=-1\,\,ta\,\,được:\\ \sin x + \cos x =1\\ \Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right) = -1\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = -\dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\end{array}\right.\qquad (k \in \Bbb Z)\\ \text{Vậy phương trình có họ nghiệm:}\,\,x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)\end{array}$