Question

cos4x+cos2x=0
√3 sin3x+cos3x=1

Answer 1

2x=a
cos(2a)+cos(a)=0
cosˆ2(a)-sinˆ2(a)+cos(a)=0
2cosˆ2(a)-1+cos(a)=0
cos(a)=x
2xˆ2+x-1=0
D=1+8=9
x=cos(a)=(-1+/-3)/4=-1; 1/2
cos(a)=-1
cos(a)=1/2
a=Pi+2*Pi*n
2x=Pi+2*Pi*n
x=Pi/2+Pi*n

cos(a)=1/2
2x=a=Pi/3+2Pi*n
x=Pi/6+Pi*n

2) sqrt(3)* sin3x+cos3x=1

2*(1/2 *sqrt(3)*sin3x + 1/2*cos3x)=1
2*(sin(pi/6)*cos3x+cos(pi/6)*sin3x)=1
2*(sin(3x+Pi/6))=1
sin(3x+Pi/6)=1/2
3x+Pi/6=Pi/6+2Pi*n
3x=2Pi*n
x=2Pi*n/3