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Điều kiện: $\left\{\begin{array}{l}\cos x\ne0\\\cos\dfrac{x}{2}\ne0\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{array}\right.,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
$\tan x+\cos x-\cos^2x=\sin x\dfrac{\cos x\cos\dfrac{x}{2}+\sin x\sin\dfrac{x}{2}}{\cos x\cos\dfrac{x}{2}}$
$\Leftrightarrow \tan x+\cos x-\cos^2x=\dfrac{\sin x\cos\dfrac{x}{2}}{\cos x\cos\dfrac{x}{2}}$
$\Leftrightarrow \tan x+\cos x-\cos^2x=\tan x$
$\Leftrightarrow \cos x-\cos^2x=0$
$\Leftrightarrow \cos x=1$
$\Leftrightarrow x=k2\pi,k\in\mathbb{Z}$
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