Solve the equation $sinx \ cosx = sqrt(2)/4$ for $0 le x le 2pi$ ?

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Answer 1 · 2017-02-11T23:59:05

$x = pi/8, (3pi)/8, (9pi)/8, (11pi)/8$

We want to solve:

$sinx \ cosx = sqrt(2)/4$

We can simplify the expression using the identity;

$sin 2A -= 2sinA \ cos A$

Which gives us:

$\ \ 1/2sin2x = sqrt(2)/4$
$:. sin2x = sqrt(2)/2$

If $0 le x le 2pi => 0 le 2x le 4pi$ , so we look for solutions of $sin theta = sqrt(2)/2$ for $theta in [0,4pi]$

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So the solutions within the range $2x in [0,4pi] \ \ (=> x in [0,2pi])$ are:

$\ \ \ \ \ 2x = pi/4, pi-pi/4, 2pi+pi/4, 3pi-pi/4$
$:. 2x = pi/4, (3pi)/4, (9pi)/4, (11pi)/4$
$:. \ \ x = pi/8, (3pi)/8, (9pi)/8, (11pi)/8$

Solve the equation sinx \ cosx = sqrt(2)/4 sinx \ cosx = sqrt(2)/4 for 0 le x le 2pi 0 le x le 2pi ?