Question

I just need some help solving this question, i’m not sure what to do

Answer 1

From the double-angle identity,

$cos2x=2*sinx*cosx$

we can rewritte our given equation as:

$4sinxcosx-2cosx=0$

By factoring 2cosx on the left hand side, we have

$2cosx(2sinx-1)=0$

This equation has 2 solutions when

$\begin{gathered} cosx=0\text{ ...\lparen A\rparen} \\ and \\ 2sinx-1=0\text{ ...\lparen B\rparen} \end{gathered}$

From equation (A), we obtain

$x=\frac{\pi}{2}\text{ or }\frac{3\pi}{2}$

and from equation (B), we have

$\begin{gathered} sinx=\frac{1}{2} \\ which\text{ gives} \\ x=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{gathered}$

On the other hand, we can find one more solution from the original equation by substituting x=0, that is,

$\begin{gathered} 2ccos(2\times0)-2cos0=0 \\ which\text{ gives} \\ 2\times1-2\times1=0 \\ so\text{ 0=0} \end{gathered}$

then, x=0 is another solution. In summary, we have obtained the following solutions:

$\begin{gathered} x=0 \\ x=\frac{\pi}{2}\text{or}\frac{3\pi}{2}\text{ and } \\ x=\frac{\pi}{6}\text{or}\frac{5\pi}{6} \end{gathered}$

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}