Đáp án: $ x=\dfrac34\pi+k\pi$ Giải thích các bước giải: Ta có: $2(\sin x+\cos x)=\dfrac12\cos2x$ $\to 2(\sin x+\cos x)=\dfrac12(\cos^2x-\sin^2x)$ $\to 2(\sin x+\cos x)=\dfrac12(\cos x-\sin x)(\cos x+\sin x)$ $\to 4(\sin x+\cos x)=(\cos x-\sin x)(\cos x+\sin x)$ $\to 4(\sin x+\cos x)-(\cos x-\sin x)(\cos x+\sin x)=0$ $\to (\sin x+\cos x)+(4-\cos x+\sin x)=0$ Vì $-1\le \sin x,\cos x\le 1$ $\to 4-\cos x+\sin x\ge 4-1-1=2>0$ $\to \sin x+\cos x=0$ $\to \sin x=-\cos x$ Nếu $\cos x=0\to \sin x=0$ vô lý vì $\sin^2x+\cos^2x=1$ $\to \cos x\ne 0$ $\to \dfrac{\sin x}{\cos x}=-1$ $\to \tan x=-1$ $\to x=\dfrac34\pi+k\pi$ Reply
$\sqrt2(\sin x+\cos x)=\dfrac{1}{2}(\cos2x)$ $\Leftrightarrow \sqrt2(\sin x+\cos x)-\dfrac{1}{2}(\cos^2x-\sin^2x)=0$ $\Leftrightarrow \sqrt2(\sin x+\cos x)+\dfrac{1}{2}(\sin x+\cos x)(\sin x-\cos x)=0$ $\Leftrightarrow (\sin x+\cos x)(\sqrt2+\dfrac{1}{2}(\sin x-\cos x))=0$ $+) \sin x+\cos x=0\Leftrightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=0\Leftrightarrow x=\dfrac{-\pi}{4}+k\pi$ $+) \sqrt2-\dfrac{1}{2}(\sin x-\cos x)=0\Leftrightarrow \sin x-\cos x=2\sqrt2\Leftrightarrow \sqrt2\sin(x-\dfrac{\pi}{4})=2\sqrt2\Leftrightarrow \sin(x-\dfrac{\pi}{4})=2>1$ (loại) Reply
Đáp án: $ x=\dfrac34\pi+k\pi$
Giải thích các bước giải:
Ta có:
$2(\sin x+\cos x)=\dfrac12\cos2x$
$\to 2(\sin x+\cos x)=\dfrac12(\cos^2x-\sin^2x)$
$\to 2(\sin x+\cos x)=\dfrac12(\cos x-\sin x)(\cos x+\sin x)$
$\to 4(\sin x+\cos x)=(\cos x-\sin x)(\cos x+\sin x)$
$\to 4(\sin x+\cos x)-(\cos x-\sin x)(\cos x+\sin x)=0$
$\to (\sin x+\cos x)+(4-\cos x+\sin x)=0$
Vì $-1\le \sin x,\cos x\le 1$
$\to 4-\cos x+\sin x\ge 4-1-1=2>0$
$\to \sin x+\cos x=0$
$\to \sin x=-\cos x$
Nếu $\cos x=0\to \sin x=0$ vô lý vì $\sin^2x+\cos^2x=1$
$\to \cos x\ne 0$
$\to \dfrac{\sin x}{\cos x}=-1$
$\to \tan x=-1$
$\to x=\dfrac34\pi+k\pi$
$\sqrt2(\sin x+\cos x)=\dfrac{1}{2}(\cos2x)$
$\Leftrightarrow \sqrt2(\sin x+\cos x)-\dfrac{1}{2}(\cos^2x-\sin^2x)=0$
$\Leftrightarrow \sqrt2(\sin x+\cos x)+\dfrac{1}{2}(\sin x+\cos x)(\sin x-\cos x)=0$
$\Leftrightarrow (\sin x+\cos x)(\sqrt2+\dfrac{1}{2}(\sin x-\cos x))=0$
$+) \sin x+\cos x=0\Leftrightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=0\Leftrightarrow x=\dfrac{-\pi}{4}+k\pi$
$+) \sqrt2-\dfrac{1}{2}(\sin x-\cos x)=0\Leftrightarrow \sin x-\cos x=2\sqrt2\Leftrightarrow \sqrt2\sin(x-\dfrac{\pi}{4})=2\sqrt2\Leftrightarrow \sin(x-\dfrac{\pi}{4})=2>1$ (loại)