Question

A. pi/3b. 2pi/3c. 4pi/3d. 5pi/3.Find the solution of each equation on the interval ( 0, 2pi).

Answer 1

We have the expression:

`2\cos x+1=0`

To find x, first, we solve for cosx by subtracting 1 to both sides of the equation:

`\begin{gathered} 2\cos x+1-1=-1 \\ 2\cos x=-1 \end{gathered}`

Then, divide both sides by 2:

`\begin{gathered} (2\cos x)/(2)=-(1)/(2) \\ \cos x=-(1)/(2) \end{gathered}`

Here we have that the value of x is such that the cosine of that value is equal to -1/2.

We can use the inverse cosine to solve for x:

`x=\cos ^(-1)(-(1)/(2))`

And the result is:

`x=2.094`

And we look in our options which one has the corresponding decimal value:

`\begin{gathered} (\pi)/(3)=1.047 \\ (2\pi)/(3)=2.094 \\ (4\pi)/(3)=4.189 \\ (5\pi)/(3)=5.236 \end{gathered}`

And we can see that 2pi/3 is equal to the value that we got for x, so:

`x=(2\pi)/(3)`

Answer:

`x=(2\pi)/(3)`