a. Sin3x + cos2x – sinx = 0 b. Sin5x + 2cos^2x = 1 c. Sinx + 4cosx = 2 + sin2x d. Cos3x + cos2x – cosx – 1 = 0 e. Sin^2 (x – π/4) = cos^2 x f. 5cosx

0 bình luận về “a. Sin3x + cos2x – sinx = 0 b. Sin5x + 2cos^2x = 1 c. Sinx + 4cosx = 2 + sin2x d. Cos3x + cos2x – cosx – 1 = 0 e. Sin^2 (x – π/4) = cos^2 x f. 5cosx”

1. a)

    sin3x +cos2x -sinx=0
   <=>(sin3x-sinx)+cos2x=0
   <=>2cos2x.sin x+cos2x=0
   cos2x(2sinx+1)=0
   cos2x=0; 2x=π/2+kπ
   x=π/4+kπ!2
   2sinx+1=0;
   x=-π/6+k2π
   x=5π/6+k2π

   b)

   sin5x + 2cos^2x = 1
   <=>sin5x =1-2cos^2x=-cos2x=cos(-2x)=sin[(π/2-(-2x)]
   =sin(2x+π/2)
   5x=2x,+π/2+k2π; x=π/6+2kπ/3
   5x=-2x+π/2+k2π; x=π/14+k2π/7
   k€Z

   c)

   sinx + 4cosx = 2 + sin2x
   <=> sinx + 4cosx = 2 + 2sin xcosx
   <=> [sinx – 2sinx cosx] + 4cosx -2 =0
   <=> sinx(1 – 2 cosx] -2(1-2cosx) =0
   <=> (1 – 2 cosx)(sinx -2) =0
   <=> (1 – 2 cosx) =0
   cosx =1/2
   x =+-pi /3 + k2pi

   d)

   cos3x+cos2x-cosx-1=0
   >>4cos^3(x)-3cosx+2cos^2(x)-1-cosx-1=0
   >>4cos^3(x)+2cos^2(x)-4cosx-2=0
   >>2cos^3(x)+cos^2(x)-2cosx-1=0
   >>cos^2(x)(2cosx+1)-(2cosx+1)=0
   >>(2cosx+1)(cos^2(x)-1)=0
   >>(2cosx+1)(cosx-1)(cosx+1)=0
   >>2cosx+1=0; cosx-1=0 hoặc cosx+1=0

   ⇔−15π8≤kπ≤π8⇔−158≤k≤18⇔−1,875≤k≤0,125(k∈Z)⇒k∈{−1;0}⇒[x=−π8x=7π8.

    

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