How do you solve the following equation $sin x cos x= 1/2$ in the interval [0, 2pi]?

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Answer 1 · 2016-10-04T15:36:45

$S={pi/6, pi/3,(5pi)/6,(5pi)/3}$

$sinxcosx=1/2$

$sinx = 1/2, cosx=1/2$

$x=sin^-1(1/2)$ OR $x=cos^-1(1/2)$

$x=pi/6 +2pin, x=(5pi)/6+2pin$ OR $x=+-pi/3 +2pin$

$n=0, x=pi/6,(5pi)/6, pi/3,-pi/3$

$n=1, x=(13pi)/6, (17pi)/6,(7pi)/3,(5pi)/3$

$S={pi/6, pi/3,(5pi)/6,(5pi)/3}$

How do you solve the following equation sin x cos x= 1/2sin x cos x= 1/2 in the interval [0, 2pi]?