From the double-angle identity,
[tex]cos2x=2*sinx*cosx[/tex]we can rewritte our given equation as:
[tex]4sinxcosx-2cosx=0[/tex]By factoring 2cosx on the left hand side, we have
[tex]2cosx(2sinx-1)=0[/tex]This equation has 2 solutions when
[tex]\begin{gathered} cosx=0\text{ ...\lparen A\rparen} \\ and \\ 2sinx-1=0\text{ ...\lparen B\rparen} \end{gathered}[/tex]From equation (A), we obtain
[tex]x=\frac{\pi}{2}\text{ or }\frac{3\pi}{2}[/tex]and from equation (B), we have
[tex]\begin{gathered} sinx=\frac{1}{2} \\ which\text{ gives} \\ x=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{gathered}[/tex]On the other hand, we can find one more solution from the original equation by substituting x=0, that is,
[tex]\begin{gathered} 2ccos(2\times0)-2cos0=0 \\ which\text{ gives} \\ 2\times1-2\times1=0 \\ so\text{ 0=0} \end{gathered}[/tex]then, x=0 is another solution. In summary, we have obtained the following solutions:
[tex]\begin{gathered} x=0 \\ x=\frac{\pi}{2}\text{or}\frac{3\pi}{2}\text{ and } \\ x=\frac{\pi}{6}\text{or}\frac{5\pi}{6} \end{gathered}[/tex]However, the intersection of the last set is empty. So the unique solution is x=0 as we can corroborate on the following picture:
Therefore, the solution set is: {0}
