# How do you find all of the exact solutions on the interval [0,2pi) for the following equation? 2cos(1/2 X) = -root3

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### 2 Answers

Solve `2cos(1/2x)=-sqrt(3)` on `[0,2pi)` :

`2cos(x/2)=-sqrt(3)`

`cos(x/2)=-sqrt(3)/2`

` `` ``x/2=(5pi)/6` or `(7pi)/6` ** `x/2=cos^(-1)(-sqrt(3)/2)` ; remember to look at all quadrants as the inverse function is defined only in quadrants I and II**

`x/2=(5pi)/6 ==>x=(5pi)/3`

`x/2=(7pi)/6 ==>x=(7pi)/3` but this is not in the domain.

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`2cos(1/2x)=-sqrt(3)` has the only solution of `x=(5pi)/3` in theinterval `[0,2pi)`

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The graph of `y=2cos(1/2x)`and `y=-sqrt(3)` :

You need to divide both sides by 2 to keep only the cosine function to the left, such that:

`cos(x/2) = -sqrt3/2 =gt x/2 = cos^(-1)(-sqrt3/2)`

You need to remember that the values of cosine function are negative in qudrants 2 and 3. Since `sqrt3/2` is the value of cosine of an angle that measures pi/6 radians, hence, the corresponding angle in quadrant 2 is `pi - pi/6` and the corresponding angle in quadrant 3 is `pi + pi/6` such that:

`x/2 = pi - pi/6 or x/2 = pi + pi/6`

`x = 2pi - 2pi/6 =gt x = 2pi - pi/3 =gt x = 5pi/3`

`x = 2pi + 2pi/6 =gt x = 2pi + pi/3 =gt x = 7pi/3`

**Hence, evaluating the exact solutions over the interval [0,2pi) yields `x = 5pi/3` and `x = 7pi/3` .**