cos(6x)=sqrt3sin(6x)
sqaure both sides
cos^2(6x) = 3sin^2(6x)
cos^2(6x) = 3 [ 1 - cos^2(6x)]
cos^2(6x) = 3 - 3cos^2(6x) re-arrange as
4cos^2(6x) - 3 = 0 factor this as
[2cos(6x) + √3 ] [ 2 cos(6x) - √3] = 0
Setting the first factor to 0 we have
2cos(6x) + √3 = 0
cos(6x) = -√3 / 2
cos(x) = -√3 / 2 at 150° + n*360° and at 210° + n*360° where n is an integer
Dividing each of these by 6, we have :
x = 25° + n* 60° and x = 35° + n* 60°
As will be shown shortly.....only the second solution is good......the first is an extraneous solution produced by squaring both sides of the original equation
Settintg the second factor to 0 , we have
2cos(6x) - √3 = 0
cos(6x) = √3 / 2
cos(x) = √3 / 2 at 30° + n * 360° and at 330° + n*360°
Dividing each of these by 6 we have :
x = 5° + n* 60° and x = 55° + n* 60°
For the same reason as above, only the first solution is good.
Here's a graph of the solutions on the interval [ 0, 360 ]°
https://www.desmos.com/calculator/ywvf0g7bxs
