Đáp án:
1. \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x=±\dfrac{\pi}{6}+k\dfrac{\pi}{2}\\x=±\pi+k\dfrac{\pi}{2}\end{array} \right.\)
2. \(\left[ \begin{array}{l}x=±\dfrac{2\pi}{3}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\end{array} \right.\)
Giải thích các bước giải:
1. phương trình ⇔ $(1+\cos 6x).\cos 2x-(1+\cos 2x)=0$
⇔ $(1+4\cos^3 2x-3\cos 2x).\cos 2x-(1+\cos 2x)=0$
⇔ $(1+\cos 2x)(\cos^2 2x+1+\cos 2x+3\cos^2 2x-3\cos 2x)-(1+\cos 2x)=0$
⇔ $(1+\cos 2x)(2\cos 2x-1).\cos 2x=0$
⇔ \(\left[ \begin{array}{l}\cos 2x=-1\\\cos 2x=0\\\cos 2x=\dfrac{1}{2}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x=±\dfrac{\pi}{6}+k\dfrac{\pi}{2}\\x=±\pi+k\dfrac{\pi}{2}\end{array} \right.\)
2. phương trình
⇔ $\sin^2 x+\cos^2 x +\sin x+\cos x +2\sin x.\cos x+\sin^2 x-\cos^2 x =0$
⇔ $2\cos^2 x+2\sin x.\cos x+\sin x+\cos x=0$
⇔ $(2\cos x+1)(\sin x+\cos x)=0$
⇔ \(\left[ \begin{array}{l}\cos x=-\dfrac{1}{2}\\\tan x=-1\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=±\dfrac{2\pi}{3}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\end{array} \right.\)
