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sorry forgot bout the othersthanx...wat bout the oda 3qs?
here:
may/june 09
4ii)
From 4i) u got ur values for a,b and c.
a=6 b=2 and c=3, right?
so u jst put these values into this equation ur given and solve for x
====> y = 6sin(2x) + 3
so when y=0,
6sin(2x) +3 = 0
sin(2x) = -1/2
since you can't solve for negative values u can say:
let sin2x = 0.5
==> 2x = pi/6 (angles are in radians)
what i do frm here i let 2x = alpha (for alpha i'll use @)
so @ = pi/6
now you find the angles for sin where it is negative that will be in the 3rd and 4th quadrant
so,
in 3rd quadrant:
@ = pi + pi/6 = 7pi/6
in 4th quadrant:
@ = 2pi - pi/6 = 11pi/6 <-----this angle u can eliminate it coz u can see it's going to be the largest angle
now u can find x:
2x = 7pi/6
so, x = 7pi/12
hence the smallest is 7pi/12
i know its kinda long but thats how i do it, hope u understood