How do you solve $sinx*cosx=1/2$ for x in the interval [0,2pi)?

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Answer 1 · 2016-01-31T05:50:25

$x = pi/4$

Use trig identity: sin 2x = 2sin x.cos x
sin x.cos x = $(1/2)sin 2x = 1/2$ --> sin 2x = 1
sin 2x = 1 --> arc $2x = pi/2$ --> $x = pi/4$

Check
sin (pi/4).cos (pi)/4 = (sqrt2/2)(sqrt2/2) = 2/4 = 1/2 . OK

How do you solve sinx*cosx=1/2sinx*cosx=1/2 for x in the interval [0,2pi)?