# cos3X+cos5X=0

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Solve `cos3x+cos5x=0` :

(1) `cos(3x)=cos(2x+x)=cos2xcosx-sin2xsinx`

`=(2cos^2x-1)cosx-2sin^2xcosx`

`=2cos^3x-cosx-2(1-cos^2x)cosx`

`=2cos^3x-cosx-2cosx+2cos^3x`

`=4cos^3x-3cosx`

(2) `cos(5x)=cos(3x+2x)`

`=cos3xcos2x-sin3xsin2x`

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** `sin3x=4cos^2xsinx-sinx`

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`=(4cos^3x-3cosx)(2cos^2x-1)-(4cos^2xsinx-sinx)(2sinxcosx)`

`=8cos^5x-10cos^3x+3cosx-[8cos^3xsin^2x-2sin^2xcosx]`

`=8cos^5x-10cos^3x+3cosx-[8cos^3x-8cos^5x-2cosx+2cos^3x]`

`=16cos^5x-20cos^3x+5cosx`

(3) `cos3x+cos5x=16cos^5x-16cos^3x+2cosx`

(4) `16cos^5x-16cos^3x+2cosx=0`

`2cosx(8cos^4x-8cos^3x+1)=0` So `2cosx=0` or:

Using the quadratic formula we get:

`cos^2x=(8+-sqrt(64-4(8)(1)))/16`

`cos^2x=1/2+-sqrt(2)/4`

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

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

`cosx=+-sqrt(2+-sqrt(2))/2`  or `2cosx=0`

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Thus `x=(npi)/2`

`cos^(-1)(sqrt(2+sqrt(2))/2)`

`cos^(-1)(-sqrt(2+sqrt(2))/2)`

`cos^(-1)(sqrt(2-sqrt(2))/2)`

`cos^(-1)(-sqrt(2-sqrt(2))/2)`

For each of the last 4 solutions, if the argument of the arccosine (inverse cosine) is k, then `cos^(-1)(2pi-k)` is also a solution. Thus there are ten solutions on each interval of `2pi` .

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The graph of `y=cos3x+cos5x` :

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`cos3x+cos5x = 0`

`2cos4x*cosx = 0`

`cos4x*cosx = 0`

For this to happen;

`cos4x = 0` OR `cosx = 0`

When `cos4x = 0`

`cos4x = 0`

`cos4x = cos(pi/2)`

`4x = 2npi+-(pi/2)` Where `n in Z`

`x = npi/2+-pi/8`

When n = -1 then `x = -3/8` OR `x = -5pi/8`

When n = 0 then `x = pi/8` OR `x = -pi/8`

When n = 1 then `x = 5pi/8` OR `x = 3pi/8`

When `cosx = 0`

`cosx = 0`

`cosx = cos(pi/2)`

` x = 2mpi+-pi/2` where `m in Z`

When m = -1 then `x = (-3pi)/2` OR `x = (-5pi)/2`

When m = 0 then `x = pi/2` OR `x = -pi/2`

When m = 1 then `x = (5pi)/2` OR `x = (3pi)/2`

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