The cosecant function #color(red)csc x# is equivalent to #color(red)(1/sin x#.
Plugging this into the equality #csc x = 15/2#, you'll get
#1/sin x = 15/2 => sin x = 2/15#.
In order to find #sin 2x#, we have to use the #color(red)("sum formula")#, that exclaims
#sin(alpha+beta) = sinalphacosbeta+cosalphasinbeta#
#sin 2x# can be written as #sin (x+x)#, so we can apply the identity above :
#sin2x = sin(x+x) = sinxcosx+cosxsinx =2sinxcosx#.
From the identity #sin^color(red)2 x + cos^color(red)2x = 1#, it follows that #cos x = +-sqrt(1-sin^color(red)2x)#.
Because #x in# quadrant 1, #cos x# is positive :
#cos x = sqrt(1-(2/15)^2) = sqrt(1-4/225) = sqrt (221/225)= sqrt(221)/sqrt(225) = sqrt(221)/15#
Finally,
#sin 2x =2*2/15*sqrt(221)/15#
#color(red)(sin2x = 4sqrt(221)/225#
I would rather not make this answer any more longer and clustered, so I highly suggest checking out this algorithm of computing square roots:
https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Decimal_(base_10)
Afterwards, you'll find #sqrt 221# to be equal to #~~14.8660#, up to #4# decimal places.
#color(red)(sin 2x ~~ 4*14.8660/225 ~~ 0.26428 = 0.2643#.
