Solve tan 2x = 8 cos^2 x - cot x , 0 <=2x<=pi

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beckden | High School Teacher | (Level 1) Educator

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`tan2x=8cos^2x-cotx`

`sin(2x)/cos(2x)=8cos^2(x)-cos(x)/sin(x)`

Multiply by sin(x)cos(2x)

`sin(2x)sin(x)=8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)`

Rearanging

`8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)-sin(2x)sin(x)=0`

Using `sin(2x)=2sin(x)cos(x)`

`8cos^2(x)sin(x)cos(2x)-cos(2x)cos(x)-2sin^2(x)cos(x)=0`

Factor out cos(x)

`cos(x)(8cos(x)sin(x)cos(2x)-cos(2x)-2sin^2(x)) = 0`

Using `2sin^2(x)=1-cos(2x)` and `cos(x)sin(x) = 1/2sin(2x)` from the double angle formulas

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

Using `2cos(2x)sin(2x)=sin(4x)`

`cos(x)(2sin(4x)-cos(2x)-1+cos(2x)) = 0`

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

so either `cos(x) = 0` or `2sin(4x)-1 = 0`

`cos(x) = 0` at `x=pi/2`

`2sin(4x)-1 = 0`

`sin(4x)=1/2` when `4x=pi/6` or `4x=(5pi)/6`

Gives us the answers `pi/2, pi/24,` and `5pi/24`

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