Question

How do you solve the equation `4tan^2u-1=tan^2u`?

Answer 1

`u = pi/6 + kpi" "`where `k in ZZ`
`" "`
Or
`u= -pi/6+kpi" "` where `k in ZZ`

Explanation:

`4tan^2u - 1 =tan^2 u`
`" "`
`rArr4tan^2u - tan^2u =1`
`" "`
`rArr3tan^2=1`
`" "`
`rArrtan^2u = 1/3`
`" "`
`rArrtanu=+1/(sqrt3) or tanu=-1/(sqrt3)`
`" "`
`rArrtanu =(sqrt3)/3 or tanu =- (sqrt3)/3`
`" "`
`u =pi/6 +kpi" "` where `pi in ZZ`
`" "`
Or
`u = -pi/6 +kpi" "`