Đáp án: $x=\dfrac{\pi }{12}+\pi k,\:x=\dfrac{11\pi }{12}+\pi k,\:x=\dfrac{5\pi }{12}+\pi k,\:x=\dfrac{7\pi }{12}+\pi k$
Giải thích các bước giải:
Ta có:
$2\cos^22x-2\sin^22x-1=0$
$\to 2\cos^22x-2(1-\cos^22x)-1=0$
$\to 2\cos^22x-2+2\cos^22x-1=0$
$\to 4\cos^22x-3=0$
$\to \cos^22x=\dfrac34$
$\to\cos2x=\pm\dfrac{\sqrt{3}}{2}$
Nếu $\cos2x=\dfrac{\sqrt{3}}{2}$
$\to 2x=\dfrac{\pi }{6}+2\pi k,\:2x=\dfrac{11\pi }{6}+2\pi k$
$\to x=\dfrac{\pi }{12}+\pi k,\:x=\dfrac{11\pi }{12}+\pi k$
Nếu $\cos2x=-\dfrac{\sqrt{3}}{2}$
$\to 2x=\dfrac{5\pi }{6}+2\pi k,\:2x=\dfrac{7\pi }{6}+2\pi k$
$\to x=\dfrac{5\pi }{12}+\pi k,\:x=\dfrac{7\pi }{12}+\pi k$
