Question #223091

Solve the following in the interval [0;2 π]

sinx > √3/2

cosx > √3/2

IcosxI ≤ 1/2


1
Expert's answer
2021-09-13T12:59:23-0400

(i)

the value of sin is +ve in first and second quadrant so the interval where sinx is positve is 0 to π0\space to\space \pi

now we want sinx to be greater then 32\frac {\sqrt3}{2} ,now in first quadrant the graph of sin increases from o to π2o\space to\space \frac{\pi}{2}

and sinπ3sin\frac{\pi}{3} is 32\frac {\sqrt3}{2} so in the interval π3 to π2\frac{\pi}{3}\space to\space \frac{\pi}{2} sin should be greater than 32\frac {\sqrt3}{2}

now in second quadrant the graph of sin decreases since the value of sinπ2sin\frac{\pi}{2} is 1 and the value of sin2π3sin\frac{2\pi}{3} is 32\frac {\sqrt3}{2} so the total interval in which sinx is greater than 32 is (π3 to 2π3)\frac {\sqrt3}{2}\space is\space (\frac{\pi}{3}\space to\space \frac{2\pi}{3})

(ii)

the value of cos is +ve in first and fourth quadrant so the interval where cosx is positve is 0 to π2 and 3π2 to 2π0\space to\space \frac{\pi}{2} \space and\space\frac{3\pi}{2}\space to\space 2\pi since cos0 is 1 and cosx is decreasing graph in first quadrant and cosπ6 is 32cos\frac{\pi}{6}\space is \space \frac {\sqrt3}{2}

so in first quadrant cosx greater than 32\frac {\sqrt3}{2} in interval 0 to π6\frac{\pi}{6}

now let's take fourth quadrant the value of cos increase in fourth quadrant since cos11π6 is 32cos\frac{11\pi}{6} \space is \space \frac {\sqrt3}{2} and cos2πcos2\pi is 1, so the interval in fourth quadrant is 11π6\frac{11\pi}{6} to 2π2\pi

so the total interval in which cosx is greater than 32 is (0 to π6)(11π6 to 2π)\frac {\sqrt3}{2}\space is\space (0\space to\space \frac{\pi}{6}) \cup (\frac{11\pi}{6}\space to \space 2\pi)

(iii)

let's take first quadrant graph of cosx is decrease in first quadrant and the value of cosπ6 is 12cos\frac{\pi}{6}\space is \space \frac{1}{2} since we need value less than equal to 12\frac{1}{2} so the required interval will be [π3[\frac{\pi}{3} , π2]\frac{\pi}{2}]

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 cos2π3=12|cos\frac{2\pi}{3}|=\frac{1}{2} so the required interval is [π2 , 2π3][\frac{\pi}{2}\space ,\space \frac{2\pi}{3}]

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 cos4π3=12|cos\frac{4\pi}{3}|=\frac{1}{2} so the required interval is [4π3 , 3π2][\frac{4\pi}{3}\space , \space \frac{3\pi}{2}]

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 cos5π3=12|cos\frac{5\pi}{3}|=\frac{1}{2} so the required interval is [3π2 , 5π3][\frac{3\pi}{2}\space ,\space \frac{5\pi}{3}]

so the total interval in which cosx is less than equal to 12 [π3 , 2π3][4π3 , 5π3]\frac{1}{2} \space [\frac{\pi}{3} \space ,\space \frac{2\pi}{3}] \cup [\frac{4\pi}{3}\space , \space \frac{5\pi}{3} ]



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