[Toán] PT lượng giác.

N

nguyenbahiep1

câu 1

$cos2x + 2cos^2x -1 + 4cos^32x-3cos2x = (\frac{1}{2}cos4x + \frac{1}{2}cos2x)cos2x+2 \\ \\ \Leftrightarrow 4cos^2x -6 + 8cos^32x-4cos2x = ( 2cos^22x-1+ cos2x)cos2x \\ \\ cos2x = t \in [-1,1] \\ \\ \Rightarrow 4t^2 -6 + 8t^3-4t = ( 2t^2-1+ t).t \Leftrightarrow 3(t-1)(2t^2+3t+2) = 0 $
 
N

nguyenbahiep1

câu 2

$dk: cosx \not = 0 \\ \\ \frac{(1-cosx)(1+cosx)}{(1-sinx)(1+sinx)} = \frac{(1-cosx)(cos^2x+cosx+1)}{(1-sinx)(sin^2x+sinx+1)} \\ \\ TH_1: cosx = 1 \\ \\ TH_2: \frac{1+cosx}{1+sinx} = \frac{cos^2x+cosx+1}{sin^2x+sinx+1} \\ \\ \Leftrightarrow (sinx-cosx)(2(sinx+cosx) + sinx.cosx + 3) = 0 \\ \\ tan x = 1 \\ \\ sinx+cosx = t $
 
You must log in or register to reply here.
Top Bottom