Question

Find all value of α in [0°, 360°) that will satisfy this equation? √2cos2α-1=0

Answer 1

`x= 22.5°, 157.5°, 202.5°, 337.5°`

Explanation:

`sqrt2cos(2x)-1=0`

Let `2x` be `u`:
`cos(u)= 1/sqrt2*sqrt2/sqrt2= sqrt2/2`

`u= pi/4+-2pin, (7pi)/4+-2pin`

Now replace `u` with `2x` and solve for `x` by dividing by `2`:

`2x= pi/4+-2pin, (7pi)/4+-2pin`

`x= pi/8+-pin, (7pi)/8+-pin` where `n ∈ Z`

`x= pi/8*(180°)/pi= 22.5°+180°(1)= 202.5°`
`x=(7pi)/8*(180°)/pi= 157.5°+180°(1)= 337.5°`

`x= 22.5°, 157.5°, 202.5°, 337.5°`

graph{sqrt2cos(2x)-1 [-10, 10, -5, 5]}