bài toán lượng giác lớp 11

N

nguyenbahiep1

[laTEX]cos^2x +cos^22x -cos^23x = 1 \\ \\ cos^2x + (2cos^2x-1)^2 - (4cos^3x -3cosx)^2 = 1 \\ \\ cos^2x + 4cos^4x -4cos^2x - (4cos^3x -3cosx)^2 = 0 \\ \\ TH_1: cos^2x =0 \\ \\ TH_2: 1 + 4cos^2x -4 - (4cos^2x-3)^2 = 0 \\ \\ cos^2x = t \\ \\ 4t-3-(4t-3)^2 =0 \\ \\ 4t-3 = 0 \Rightarrow cos^2x = \frac{3}{4} \\ \\ 4t-3 = 1 \Rightarrow t = 1 \Rightarrow cos^2x = 1[/laTEX]
 
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