Giải phương trình:
4/ $8x.(2x^{2}-1).(8x^{4}-8x^{2}+1) = 1$
Đặt $x=\cos\alpha$
$8\cos\alpha(2\cos^2\alpha-1)(8\cos^4\alpha-8\cos^2\alpha+1)=1$
$\iff 8\cos\alpha.\cos2\alpha.[8\cos^2\alpha(\cos^2\alpha-1)+1]=1$
$\iff 8\cos\alpha.\cos2\alpha.(-8\cos^2\alpha.\sin^2\alpha+1)=1$
$\iff 8\cos\alpha.\cos2\alpha(-2\sin^22\alpha+1)=1$
$\iff 8\cos\alpha.\cos2\alpha.\cos4\alpha=1$
$\iff 8\sin\alpha.\cos\alpha.\cos2\alpha.\cos4\alpha=\sin\alpha$
$\iff \sin8\alpha=\sin\alpha$
$\iff 8\alpha=\alpha+k2\pi$ v $8\alpha=\pi-\alpha+k2\pi$
$\iff \alpha=\dfrac{k2\pi}{7}$ v $\alpha=\dfrac{\pi+k2\pi}{9}$
Hình như đề bài thiếu điều kiện của $x \in$ [$-1;1$]