Question

cos(9x)=1/2 how to get cosx by itself

Answer 1

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
cos(x+8x)=cos(x)cos(8x)-sin(x)sin(8x)
cos(4x+4x)=cos(4x)cos(4x)-sin(4x)sin(4x)
sin(4x+4x)=2sin(4x)cos(4x)
continue simplifying until cos(9x) is left in terms of cos(x), sin(x)
then use the identity sin(x)^2+cos(x)^2=1

Answer 2

or solve for x and plug into cos(x)
cos(9x)=1/2
9x={cos^-1(1/2)+2*pi*C,-cos^-1(1/2)+2*pi*C}
9x={pi/3+2*pi*C,-pi/3+2*pi*C}
cos(x)={cos(pi/27+2/9*pi*C),cos(-pi/27+2/9*pi*C)}

Answer 3

confusing