# If 4sin^2x=1,find the value of 2+3cos^2x/1-2sin^2x?

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Given (4\sin^2x = 1), we can solve for (\sin^2x) by dividing both sides by 4:

[ \sin^2x = \frac{1}{4} ]

Now, we know that ( \cos^2x = 1 - \sin^2x ). Substituting the value of ( \sin^2x ), we get:

[ \cos^2x = 1 - \frac{1}{4} = \frac{3}{4} ]

Now, we can substitute the values of ( \sin^2x ) and ( \cos^2x ) into the expression ( \frac{2 + 3\cos^2x}{1 - 2\sin^2x} ):

[ \frac{2 + 3 \cdot \frac{3}{4}}{1 - 2 \cdot \frac{1}{4}} ] [ = \frac{2 + \frac{9}{4}}{1 - \frac{1}{2}} ] [ = \frac{\frac{8}{4} + \frac{9}{4}}{\frac{2}{2} - \frac{1}{2}} ] [ = \frac{\frac{17}{4}}{\frac{1}{2}} ] [ = \frac{17}{4} \times 2 ] [ = \frac{17}{2} ]

So, the value of ( \frac{2 + 3\cos^2x}{1 - 2\sin^2x} ) is ( \frac{17}{2} ).

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