# How do we solve the equation `cos(x)*cos(2x)=1/4` in [0, 90) ?

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### 1 Answer

Solve `cos(x)cos(2x)=1/4` on the interval `[0,90^@)` :

`cos(x)cos(2x)=1/4`

`cos(x)cos(2x)-1/4=0`

`cosx(2cos^2x-1)-1/4=0`

`2cos^3x-cosx-1/4=0`

`1/4[8cos^3x-4cosx-1]=0`

`1/4[8cos^3x-4cos^2x-2cosx+4cos^2x-2cosx-1]=0`

** Add and subtract `4cos^2x` and write `-4cosx` as `-2cosx-2cosx` **

`1/4[(2cosx+1)(4cos^2x-2cosx-1)]=0`

By the zero product property one of the three factors is zero.

`1/4!=0`

`2cosx+1=0==>cosx=-1/2` which is not true on `[0,90^@)`

`4cos^2x-2cosx-1=0` This is a quadratic in cosx; using the quadratic formula we get:

`cosx=(2+-sqrt(4-4(4)(-1)))/(2(4))`

`cosx=(2+-sqrt(20))/8=1/4+-sqrt(5)/4`

`1/4-sqrt(5)/4<0` and this cannot be on the given interval, so

`cosx=(1+sqrt(5))/4`

Then `x=cos^(-1)((1+sqrt(5))/4)=36^@`

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**The solution to `cosxcos2x=1/4` on the interval `[0,90^@)` is `x=36^@` **

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The graph of `y=cos(x)cos(2x)` and `y=1/4` :

The units are in radians -- `36^@` is equivalent to `(pi)/5` radians; `pi/5~~.628`

The intercept feature of a graphing utility could also be used.

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