Question

Giải phương trình:

\(\cos x.\cos3x-\sin2x.\sin6x-\sin4x.\sin6x=0\).

Answer 1

Answer 2

Answer 3

x=π/2+kπ

x=π/18+kπ/9

Answer 4

 

 

Answer 5

  

Answer 6

Answer 7

 

Answer 8

  

Answer 9

 

Answer 10

 

Answer 11

x = π/2 + kπ

x = π/18 + kπ/9

Answer 12

x=π/2+kπ

x=π/18+kπ/9

 

Answer 13

1/2(cos4x+cos2x)+1/2(cos8x−cos4x)+1/2(cos10x−cos2x)=0⇔1/2(cos4x+cos2x)+1/2(cos8x−cos4x)+1/2(cos10x−cos2x)=0

⇔1/2cos4x+1/2cos2x+1/2cos8x−1/2cos4x)+1/2cos10x−1/2cos2x=0⇔1/2cos4x+1/2cos2x+1/2cos8x−1/2cos4x)+1/2cos10x−1/2cos2x=0

⇔1/2cos8x+1/2cos10x=0⇔1/2cos8x+1/2cos10x=0

⇔1/2(cos8x+cos10x)=0⇔1/2(cos8x+cos10x)=0

$\iff cos9x.cosx=0$

$TH1:cos9x=0$

9x=π/2+kπ9x=π/2+kπ

x=π/18+kπ/9x=π/18+kπ/9

$TH2:cosx=0$

x=π/2+kπ

Answer 14

x = π/18 + kπ/9 , π/2 + kπ , kϵz

Answer 15

π/2+kπ

π/18+kπ9

Answer 16

Answer 17

x=π/2+kπ

x=π/18 +kπ/9