Question

Question 9 Find all solutions to the equation in the interval [0, 2n). sin 2x - sin 4x = 0 Your answer: O O 51 71 I, 31 111 z' ' ä 'ö' ' ő 31 1171 Oo, ma Clear answer

Answer 1

19 Sinza - sinha=0 sin 20 = 2 Sinocobo - Sinax-2 Sinza cosa=0 =) =) sinza [1-2005 2n] =0 sinzu=o cr 1-2C032X=0 => 2 Coszn= 1 =) Cobzn = 1/2 It sinza=0 = sin(0) Principle angle d=0 General solution 2a= ni+(-)" (0) nant :*= , T, 37.

It cos2x = 1 -cos () Principle angle d=T13 General solution 2x= 20 I 13 X=nT I J &n 2= 7 T-I, TAI , 291-7 Sinza - sinun=o it x=0, T I, 51, I, 77,31 T XE[0, 21) (10) Cosun-coan=0 (820 = 2008 -1 =) zcoszn - 1 - (032x=0 =) 2005²x- co52x-1=0 -) 2005 2-200321 +(682x-1=0 =) 2005 in (22032x-2) + 1 (ed 20-1)=0 =) (20082x+1) (C2S24-1) = 6 200n+1=0 kr) (2x-1=0 (082x=-12 con con=1 It cos2x = -² = cos( 21 ) ..2X= 2n1 I 21 a=nT I J .x= I TI-I, TTI, 25 - This

1= , 3, 4, 5, It cabana1 = cos (0) 2=0 .: 20= 2nT to x=nT x=0, TI, E [0, 21) -:x= 0, 7, 2I , II, 41, St Cosant sinn = 1-2sin 31+ sinn (.cas20 = 1-2 since) (12) we have acoso = 1 +oso costo - 1 teasca => e89 , It case ItCso 2 It cos co (2) 1+(59) It cos(57) (it cos (1-7) ( +(- CES TIG) - 110 + (-COST/6)

5) = 2