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

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embizze | High School Teacher | (Level 1) Educator Emeritus

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







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


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


`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:



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


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


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


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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