[Toán 11] Giải phương trình lượng giác: $\cos ^4x+\sin ^6x = \cos 2x$

N

nguyenbahiep1

[laTEX]\frac{(1+cos2x)^2}{4} + \frac{(1-cos2x)^3}{8} - cos2x = 0 \\ \\ cos2x = u \in [-1, 1] \\ \\ (3-u)(u-1)^2 =0 \Rightarrow cos2x = 1 \Rightarrow ?[/laTEX]
 
C

connhikhuc

ta có:

[TEX](1-sin^2 x)^2 + sin^6 x = 1-2sin^2 x[/TEX]

\Leftrightarrow [TEX]1-2sin^2 x + sin^4 x + sin^6 x = 1-2sin^2 x[/TEX]

\Leftrightarrow [TEX]sin^4 x.(1+sin^2 x) = 0[/TEX]

tự giải
 
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