A few days ago
Anonymous

can anyone plz kindly draw the solution on a paint or watever and post to imageshack. so that i will be able to see the solution.

1)sketch the graph y = – I 2 sin x I for 0 < or equal x < or equal to 2pie 2) on the same axes, sketch a suitable straight line so that the equation k + I 2 sin x I = 0 has only two solution.state the equation of the straight line 3)find the range of values for k,if k + I 2 sin x I = 0 has four solutions for 0< or equal x < or equal 2 pie

A few days ago
David F

#1. 0 —————

-1 | …….|………|

-2 \——/\——/

where the x axis goes from 0 to 2 pi.

If k + | 2 sin x | = 0, then k must equal – |2 sin x|, the graph in #1.

A straight line that touches k only twice must pass through (pi/2, -2) and (3 pi / 2, -2) [ie, the bottom of the two curves]

y = -2

#3. shouldn’t there be a ‘y has 4 solutions’ ??? Y can range from y>-2 to y < 0. If y = -2, there are 2 solutions. If y = 0, there are 3 solutions. everywhere in between are 4 solutions. If you have Microsoft Excel, you can create a table of values, and get it to graph it for you: try a1 = 0 a2 = a1 +2 * PI () / 10 a3 = a2 + 2 * pi() / 10 etc... to a 11 b1 = - abs ( 2 * sin ( a1 ) ) b2 = -abs ( 2 * sin ( a2 )) etc. to b11 then use the chart wizard to do an x-y plot from a1 to b11

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