Question

How do you solve `2cos^2x-2sin^2x=1` and find all solutions in the interval `0<=x<360`?

Answer 1

`x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6`
Where, `0<=x<360^0=>x in[0,2pi)`

Explanation:

`2cos^2x-2sin^2x=1=>cos^2x-sin^2x=1/2`
`=>cos2x=1/2=>cos2x=cos(pi/3)`
`=>2x=2kpi+-pi/3,kinZ`
`color(red)(=>x=kpi+-pi/6,kinZ)`
`kinZ=>k=0,+-1.+-2,+-3,..etc`
`k=0=>color(red)(x=pi/6in[0,2pi))`
`k=1=>color(red)(x=(5pi)/6in[0,2pi)orx=(7pi)/6in[0,2pi)`
`k=2=>color(red)(x=(11pi)/6in[0,2pi))orx=(13pi)/6!in[0,2pi)`...
`AAkinZ,k>3=>x!in[0,2pi)`
`k=-1=>x=(-5pi)/6!in[0,2pi)orx=(-7pi)/6!in[0,2pi)`..
`AAkinZ,k<-1=>x!in[0,2pi)`
Hence, `x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6`