determine general solution in radians for cosxcos2x - sinxsin2x = 0

Question

A) Determine the general solutioninradians for: cos(x)cos(2x) - sin(x)sin(2x) = 0

B) State the solutionfor xE (-pi, pi)

Answer 1

Cos(2x)cos(x)-sin(2x)sin(x)=0 
cos(2x+x) = 0 ' Note that CosACosB-SinASinB = cos(A+B) 
cos(3x) = 0 

cos(pi/2) = 0 
i) 3x = pi/2 => x=pi/6 
ii) 3x = 3pi/2 => x=3pi/6=> x= pi/2

Answer 2

cosx cos(2x) -sinxsin(2x) = 0

cos(2x) = soc^2(x) - sin^2(x) and sin(2x) = 2sinxcosx

cosx(cos^2(x) - sin^2(x) ) - 2sinxsinxcosx = 0

cosx(cos^2x - sin^2x - 2sin^2x)= 0

cosx(cos^2x -3sin^2x) = 0

cos^2x = 1 - sin^2x

cosx(1-sin^2x -3sin^2x) = 0

cosx( 1 - 4sin^2x) = 0

cosx = 0 and 1-4sin^2x = 0

cos(p/2) = 0 and sin^2x =1/4

x = p/2 and sinx = 1/2

x=p/2 and sin(p/6) = 1/2

x = p/2 and x = p/6

p/2 = 90 degree and p/6 = 30 degree