Giải pt ^^

N

nguyenbahiep1

[TEX]6cos2x - 1 = \frac{6.(1-cos2x)}{1+cos2x)} \\ 6cos^22x +5cos2x -1 = 6 - 6cos2x \\ 6cos^22x +11cos2x -7 = 0 \\ cos2x = \frac{1}{2}[/TEX]

câu 2 viết lại đề đi
 
Y

youaremysoul

1,
đk:.................
pt \Leftrightarrow $6(cos^2x - sin^2x) - sin^2x - cos^2x = 6\dfrac{sin^2x}{cos^2x}$

\Leftrightarrow $5cos^2x - 7sin^2x = 6\dfrac{sin^2x}{cos^2x}$

\Leftrightarrow $5cos^4x - 7sin^2xcos^2x = 6sin^2x$

\Leftrightarrow $5cos^4x - 7sin^2xcos^2x = 6sin^2x(cos^2x + sin^2x)$

\Leftrightarrow $5cos^4x - 13sin^2xcos^2 - 6sin^4x =0$

\Leftrightarrow $-6tan^4x - 13tan^2x + 5 = 0$


 
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