Giải hộ 8cos^4*x - 4cos2x + sin4x - 4 = 0

8cos$^{4}$x-4cos2x+sin4x-4=0

4(4cos$^{4}$-1)-4cos2x+2sin2x.cos2x=0

4(2cos$^{2}$-1)(2cos$^{2}$+1)-4cos2x+4sinx.cosx.cos2x=0

cos2x(2cos$^{2}$+1-1+sinxcosx)=0

cos2x.cosx.(2cosx+sinx)=0

cos2x=0 <-> 2x=$\frac{\pi}{2}$ + k$\pi$ <-> x=$\frac{\pi}{4}$ + k$\frac{\pi}{2}$

hoặc cosx=0 <-> x=$\frac{\pi}{2}$ + k$\pi$

hoặc 2cosx+sinx=0

<-> $\frac{2}{\sqrt[]{5}}$ cosx+ $\frac{1}{\sqrt[]{5}}$sinx=0

đặt $\frac{2}{\sqrt[]{5}}$=sina -> $\frac{1}{\sqrt[]{5}}=cosa

-> sina.cosx+sinx.cosa=0 <-> sin(x+a)=0 <-> x+a=k$\pi$ <-> x=-a+k$\pi$