Cos4x+cos2x=0 √3 sin3x+cos3x=1

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

Answer 2

4x = 1 - 2 sin^2 2x ;
 4 sin^4 2x + 3(1 - 2 sin^2 2x) - 1 = 0;
4 sin^4 2x + 3 - 6 sin^ 2x  - 1 = 0;
4 sin^4 2x - 6 sin^2 2x + 2 = 0;
2 sin^4 2x - 3 sin^2 2x + 1 = 0;
sin^2 2x = t;      - 1 ≤  t  ≤ 1;
2 t^2 - 3 t + 1 = 0;
t1 = 1; ⇒ sin^2 2x = 1;  ⇒ sin 2x = + - 1;⇒ 2x = pi/2 + pi*k;⇒    x = pi + 2 pi*k.
t2 = 1/2;   ⇒ sin^2 2x = 1/2; ⇒sin 2x = + - sgrt2/2; 2x = pi/4 + pik/2; ⇒x = pi/2 + pik; k-Z .

б) x = pi; 3 pi/2.