[toán 11]phương trình lượng giác

N

nguyenbahiep1

[TEX]tan^2x.cot^22x.cot3x = (tanx-cot2x)(tanx+cot2x) +cot3x \\ \frac{sin^2x}{cos^2x}.\frac{cos^22x}{sin^2x}.\frac{cos3x}{sin 3x } = \frac{-cos3x.cosx}{cos^2x.sin^22x} + \frac{cos3x}{sin 3x } \\ TH_1 : cos3x = 0 \\ \frac{sin^2x}{cos^22x}. \frac{cos^22x}{sin^2x}. \frac{1}{sin 3x} = \frac{-cosx}{cos^2x.sin^22x} + \frac{1}{sin 3x } \\ sin^2x.cos^22x = -cosx.sin3x + cos^2x.sin^22x \\ cosx.sin3x = (cosx.sin2x - sinx.cos2x).(cosx.sin2x + sinx.cos2x) \\ cosx.sin 3x = sinx.sin3x \\ \Rightarrow sin 3x = 0 \\ \Rightarrow cosx = sin x \Rightarrow tan x = 1 [/TEX]


điều kiện tự làm nhé
 
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