[Toán 11 ]Pt lượng giác

S

saotoivanyeu_nd

(sinx)^2 * (sin x +1 ) + cosx(cox+1)=0
(1 - cos^2 x)* (sin x +1 ) + cosx(cox+1)=0
(1 - cosx)(1+cosx)* (sin x +1 ) + cosx(cox+1)=0
nhom 1+cosx vao la ra luon do .Nho tk toi do hj tuy bai nay de
 
N

ngoctuan123

[tex]sin^3x[/tex] + cos2x+cosx =0
<=> [tex]sin^3x[/tex] +[tex]cos^2x[/tex] -[tex]sin^2x[/tex] +cosx=0
<=> [tex]sin^2x[/tex] (sinx-1)+cosx(cosx+1)=0
<=> (1-[tex]cos^2x[/tex] )(sinx-1)+cosx(cosx+1)=0
<=> (cosx+1)(cosx+(sinx+1)(1-cosx))=0
<=> [tex]\left\{ \begin{array}{l} cosx = -1 \\ 1-sinxcossx+sinx=0 \end{array} \right.[/tex]
bạn tự làm tiếp nha
 
C

connguoivietnam

sin^3x + cos2x + cosx=0
sinx.sin^2x + 2cos^2x-1 + cosx=0
sinx.(1-cos^2x) + 2cos^2x+cosx -1=0
sinx.(1-cosx)(1+cosx)+(cosx+1)(2cosx-1)=0
(1+cosx)[sinx(1-cosx)+(2cosx-1)]=0
cosx=-1
sinx(1-cosx)+(2cosx-1)=0
sinx-sinxcosx+2cosx-1=0
cosx+cosx-sinxcosx-1+sinx=0
cosx+cosx(1-sinx)-(1-sinx)=0
cosx+(1-sinx)(cosx-1)=0
[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]+[cos(x/2)-sin(x/2)]^2.(cosx-1)=0
[cos(x/2)-sin(x/2)][cos(x/2)+sin(x/2)+[cos(x/2)-sin(x/2)].(cosx-1)]=0
 
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