Question

How do you solve `6sin^2x+5=8` for `0<=x<=pi`?

Answer 1

The solution is `x=pi/4, x=(3pi)/4` in interval `0<=x<=pi`

Explanation:

`6sin^2x+5=8 or 6sin^2x=3 or sin^2x=1/2or sinx=+-1/sqrt2`
In interval `0<=x<=pi ; sin (pi/4) =1/sqrt2 , sin(pi-(pi)/4)=1/sqrt2` and
`sin (pi+pi/4)= -1/sqrt2` But `x=(pi+pi/4)` is outside the given interval.
Hence the solution is `x=pi/4, x=(3pi)/4` in interval `0<=x<=pi`[Ans]