[Toán 11] Phương trình lượng giác

N

niemkieuloveahbu

[TEX]PT \Leftrightarrow (1-2sin^2x)sin^8x-(2cos^2x-1)cos^8x=\frac{5}{4}cos2x\\ (sin^8x-cos^8x)cos2x=\frac{5}{4}cos2x\\ \Leftrightarrow \[cos2x=0\\ sin^8x-cos^8x=\frac{5}{4}(VN) \Rightarrow x=\frac{\pi}{4}+k\frac{\pi}{2},k \in Z[/TEX]
 
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