# solve cos5x = 1/2i have tried using (cos2x)(cos2x)(cosx) = 1/2 but i dont know how to fit the 1/2 in.

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Solve `cos5x=1/2` :

There are a couple of methods. Here are 2 that I don't recommend:

(1)Rewrite `cos5x` as an angle sum:

`cos5x=cos(4x+x)=cos4xcosx-sin4xsinx`

`=(cos2(2x))cosx-(sin2(2x))sinx`

`=(cos^2(2x)-sin^2(2x))cosx-2sin2xcos2xsinx`

... Continuing to use angle sum formulas.

(2) Using the cosine of multiple angles identity:

`cosntheta=sum_(k=0)^n([n],[k])cos^kthetasin^(n-k)thetacos(1/2(n-k)pi)`

This also seems unappetizing.

(3) How about this:

`cos5theta=1/2` . The period for the graph of `y=cos5x` is `p=(2pi)/5` . We can find the points on `[0,10pi)` where `cosx=1/2` then divide the x-values by 5 to get the points on `[0,2pi)` where `cos5x=1/2` .

We get: `cos5x=1/2==>x=pi/15,pi/3,(7pi)/15,(11pi)/15,...`

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**The solutions to `cos5x=1/2` are `pi/15+(2pi)/5k` and `pi/3+(2pi)/5k` where k is an integer.**

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The graph (x values marked in `pi/15` increments): ((showing the graph of y=cos(5x) and y=1/2 to show the intersections))

A larger graph: