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Tanvi is that 5cos(27-0.3)
sorry, 2y - 0.3, not 27 -_-
hhmm - i dont recall solving one like this before - i'll have to check my list of trig identities - wait a few minutes
sure, suree, take your time. even if we replace (2y - 0.3) by a, then we would get 4sin(a) + 5cos(a) = 0. oh damn, i got it!!
tats the way to go
i don't know what that means, but it'll become tan (a) = -5/4, and then i'll solve it.
yes - thats an easier way - do you really need anyone to help you tanvidals??!! lol well done
yes, can you give me another value for a??? apart from -0.09 rad?? it needs to be positive.
tan -5/4 - the possible values are between o and 2pi radians are in the second and 4th quadrant so we get pi - 0.9 and 2pi - 0.9
i need between 0 and pi.
yes so a = 2.24 , 5.38 so 2y - 0.3 = 2.24 y = 2.54 / 2 = 1.27 rad a=5.38 will be too big one value for y = 1.27 rad
i needed to go higher than pi for value of a because y must have value 0 to pi
thank youu :)
yw - but i think ill have another coffee to stimulate my brain a bit!
oh - plz click on good answer to close the question if you are happy with the answer
i've just checked the answer on wolfram alpha - it gives it as 1.2727
thats a good site for verification
run your cursor over the root plot - the red spot on positive x-axis - i gives you this value