# `cos(2x)(2cos(x) + 1) = 0` Solve the equation.

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### 1 Answer

`cos(2x)(2cos(x)+1)=0`

solving each part,

cos(2x)=0

General solutions for cos(2x)=0 are,

`2x=pi/2+2pin , (3pi)/2+2pin`

`x=(pi+4pin)/4 , (3pi+4pin)/4`

Solving 2cos(x)+1=0,

`2cos(x)=-1`

`cos(x)=-1/2`

General solutions for cos(x)=-1/2 are,

`x=(2pi)/3+2pin , (4pi)/3+2pin`

So the solutions are,

`x=(2pi)/3+2pin , (pi+4pin)/4 , (3pi+4pin)/4 , (4pi)/3+2pin`