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

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1. 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$\pi$

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