Question

How do you find the exact value of `cos2theta-cos^2theta+costheta+1/4=0` in the interval `0<=theta<2pi`?

Answer 1

`color(blue)(pi/3 , (5pi)/3`

Explanation:

Identity:

`color(red)bb(cos(2x)=2cos^2(x)-1)`

Substitute this into the equation:

`2cos^2(theta)-1-cos^2(theta)+cos(theta)+1/4=0`

Simplify:

`cos^2(theta)+cos(theta)-3/4=0`

`4cos^2(theta)+4cos(theta)-3=0`

Let `\ \ \ \ u=cos(theta)`

Then:

`4u^2+4u-3=0`

Factor:

`(2u+3)(2u-1)=0=>u=-3/2 and u=1/2`

But:

`u=cos(theta)`

`cos(theta)=1/2`

`cos(theta)=-3/2` *

• This has no real solutions, because:

`color(red)(-1<=cos(theta)<=1`

`cos(theta)=1/2`

`theta=arccos(cos(theta))=arccos(1/2)=>theta=color(blue)(pi/3 , (5pi)/3`