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=(\pi)/(2)\text{ or }(3\pi)/(2)`

and from equation (B), we have

`\begin{gathered} sinx=(1)/(2) \\ which\text{ gives} \\ x=(\pi)/(6)\text{ or }(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*0)-2cos0=0 \\ which\text{ gives} \\ 2*1-2*1=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=(\pi)/(2)\text{or}(3\pi)/(2)\text{ and } \\ x=(\pi)/(6)\text{or}(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}