Question

Giải phương trình:

\(\sin2x-2\sin^2x=2\cos2x\).

 

Answer 1

Sin 2x-2Sin2X=2có2X

Answer 2

Answer 3

Answer 4

X=π/4+kπ

x=π/2+kπ

Answer 5

 

 

Answer 6

   

Answer 7

 

Answer 8

  

Answer 9

 

Answer 10

 

Answer 11

x = π/4 + kπ

x = π/2 + kπ

Answer 12

x=π/4+kπ

x=π/2+kπ

Answer 13

x = π/2 +kπ , π/4 +kπ , kϵz

Answer 14

\(\pi\)/4+k\(\pi\)

Answer 15

Answer 16

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Answer 17

x=π/4+kπ

x=π/2 +kπ

Answer 18

sin2x−2sin2x=2cos2x

\Leftrightarrow2\sin^2x-2\sin x\cos x+4\cos^2x-2=0⇔2sin2x−2sinxcosx+4cos2x−2=0

Xét  $\cos x=0$, phương trình thỏa mãn.

Suy ra x=\dfrac{\pi}{2}+k\pi\left(k\in\mathbb{Z}\right)x=2π​+kπ(k∈Z) là nghiệm của phương trình.

Xét $\cos x\ne 0$, chia cả hai vế của phương trình cho \cos^2xcos2x ta được:

2\tan^2x-2\tan x-2\left(1+\tan^2x\right)+4=02tan2x−2tanx−2(1+tan2x)+4=0

$\iff$\tan x=1\Leftrightarrow x=\dfrac{\pi}{4}+k\pitanx=1⇔x=4π​+kπ

Vậy các nghiệm của phương trình là x=\dfrac{\pi}{4}+k\pix=4π​+kπ và x=\dfrac{\pi}{2}+k\pi\left(k\in\mathbb{Z}\right)x=2π​+kπ(k∈Z)

Answer 19

a) sin2x - 2sin²x = 2cos2x
<=>2sinxcosx -2sin²x=2.(1-2sin²x)
<=>2sinxcosx=2-2sin²x
sinxcosx=cos²x
cosx(sinx-cosx)=0
√2.cosx.sin(x-π/4)=0
cosx=0; x=π/2+kπ
sin(x-π/4)=0; x-π/4=kπ
x=π/4+kπ

Answer 20

sin2x=2(1-2sin²x)+2sin²x
<->2sinxcosx=2-2sin²x
<->sinxcosx=cos²x
<->sinx=cosx
<->sinx=sin(π/2 -x)
=>x=π/2 -x+k2π và x=π-π/2-x+k2π
=>x=π/4+kπ