Question

Ma poate ajuta careva la exercitiul 4? Multumesc anticipat! :)

Answer 1

(a) sin3x=sin5x il scriem pe sin5x=sin(3x+2x)
sin3x=sin(3x+2x)
sin3x-sin(3x+2x) = 0
folosim formula produsului  : sina-sinb=2sin(a-b)/2*cos(a+b)/2
2sin[3x-(3x+2x)]/2 * cos[3x+(3x+2x)]/2=0
2sin(-x)*cos4x=0
-2sinx*cos4x=0
-2six*cos(2x+2x)=0      folosim formula : cos(a+b)=cosa*cosb - sina*sinb
-2sinx*[cos^2 (2x) - sin^2 (2x)]=0   ,  unde 2x=t si folosim  formula : sin^a+cos^2a = 1 acum pt sin^2(t)+cos^2(t) = 1  => ca                               sin^2 (t)=1 - cos^2 (t)
avem :
-2sinx*(cos^2(t)-1+cos^2(t))=0
-2sinx=0
sinx=0 unde x=0 si x=180
si
cos^2(t)-1+cos^2(t) = 0
2*cos^2(t)=1
cos^2(t)=1/2
cost=$+\sqrt{1/2}$
cost=$-\sqrt{1/2}$
avem insa 2x=t    unde t = + radical(1/2)=(radical2)/2
                                       t=- radical(1/2)  =-(radical2)/2
 2x=+(radical2)/2
4x=+(radical2)      x=+(radical2)/4

2x=-(radical2)/2
4x=-(radical2)    x=- (radical2)/4