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

1 Answer
Binayaka C.
Sep 30, 2016

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]

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