câu1
tự làm TXĐ nhé
[TEX]cos^2x-1=\frac{cos2x}{1+tanx}+sin^2x-\frac{1}{2}.sin2x \\ -sin^2x = \frac{(cosx-sinx)(cosx +sinx)cosx}{sinx +cosx} + sin^2x - sinx.cosx \\ (cosx-sinx)cosx + 2.sin^2x -sinxcosx = 0 \\ cos^2x + 2sin^2x -2sinxcosx = 0 \\ 2tan^2x -2tanx + 1 = 0 (V/N)[/TEX]
câu 2
[TEX]sin^2( \frac{x}{2} - \frac{\pi}{4}) .tan^2x - cos^2\frac{x}{2} = 0 \\ ( 1 - cos(x -\frac{\pi}{2})).tan^2x - 1 + cosx = 0 \\ \frac{(1-sinx)sin^2x}{cos^2x} - (1-cosx) = 0 \\ \frac{(1-sinx)(1-cosx)(1+cosx)}{(1-sinx)(1+sinx)} - (1-cosx) = 0 \\ \Rightarrow cosx = 1\\ \frac{((1+cosx)}{(1+sinx)} - 1 = 0 \\ 1 + cosx = 1 +sinx \Rightarrow tan x = 1[/TEX]
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