Express 6cos(2x) + sin(x) in terms of sin(x), hence solve the equation 6cos(2x) + sin(x) = 0 for 0<x<360

For the 1st part of the question: use the double angle formula to rewrite cos(2x) = cos^2(x) - sin^2(x). Then use the basic identity to write cos^2(x) = 1-sin^2(x), hence cos(2x) = 1-2sin^2(x). Plug the rewritten version of cos(2x) into the equation would give us 6-12sin^2(x)+sin(x).// For the 2nd part: 6-12sin^2(x)+sin(x)=0, multiply through by -1 to make the highest power coefficient positive for simplicity: 12sin^2(x)-sin(x)-6=0. Then let y=sin(x), that give us a nice quadratic 12y^2-y-6=0 which we can simply solve: (3y+2)(4y-3)=0, y=-2/3 y=3/4. For y=sin(x)=-2/3, x=-41.8 but we want x between 0-360 degrees. One way to go around that is to draw a sin graph, mark where -41.8degree is, draw a horizontal line to see where it cross the sin wave for 0<x<360, it would give that x=221degree and 318degree. Then for the other root sin(x)=3/4, x=48.6, mark it on the sin graph and draw a horizontal line. This time 48.6degree is within 0 to 360, and the line also cross when x=131degree. Hence answers are x= 48.6, 131, 221 and 318.