Solve the following in the interval [0;2 π]
sinx > √3/2
cosx > √3/2
IcosxI ≤ 1/2
(i)
the value of sin is +ve in first and second quadrant so the interval where sinx is positve is
now we want sinx to be greater then ,now in first quadrant the graph of sin increases from
and is so in the interval sin should be greater than
now in second quadrant the graph of sin decreases since the value of is 1 and the value of is so the total interval in which sinx is greater than
(ii)
the value of cos is +ve in first and fourth quadrant so the interval where cosx is positve is since cos0 is 1 and cosx is decreasing graph in first quadrant and
so in first quadrant cosx greater than in interval 0 to
now let's take fourth quadrant the value of cos increase in fourth quadrant since and is 1, so the interval in fourth quadrant is to
so the total interval in which cosx is greater than
(iii)
let's take first quadrant graph of cosx is decrease in first quadrant and the value of since we need value less than equal to so the required interval will be ,
Now let's move forward to second quadrant the graph of cosx is still decrease in second quadrant but we have |cosx| the graph will be increasing the value of so the required interval is
Now let's move forward to 3rd quadrant, graph of cosx increase is 3rd quadrant but we have |cosx| so the graph will be decreasing and the value of so the required interval is
Now let's move forward to 4th quadrant in this the value of cosx is positive so |cosx| will not effect so the value of so the required interval is
so the total interval in which cosx is less than equal to
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