Solution:
Consider the following trigonometric equation:
[tex]\sqrt[]{3}+2\cos (2u)=0[/tex]
to solve this equation, subtract the root of 3 from both sides of the equation, to obtain:
[tex]2\cos (2u)=-\sqrt[]{3}[/tex]
now, solving for cos(2u), we get:
[tex]\cos (2u)=-\frac{\sqrt[]{3}}{2}[/tex]
now, by the trigonometric circle, the general solutions for the above equation are:
[tex]2u\text{ = }\frac{5\pi}{6}+2\pi n\text{ , 2u = }\frac{7\pi}{6}+2\pi n\text{ }[/tex]
now, if we solve the above equations for u, we get the following general solutions:
[tex]u\text{ = }\frac{5\pi}{12}+\pi n\text{ , u = }\frac{7\pi}{12}+\pi n\text{ }[/tex]
so, for the interval [0,2pi), the solution would be:
[tex]u\text{ = }\frac{5\pi}{12},\text{ u=}\frac{7\pi}{12},\text{ u=}\frac{17\pi}{12},\text{ u=}\frac{19\pi}{12}[/tex]
