How do you solve $sin x \cdot cos x = \frac{1}{2}$ $sinx*cosx=1/2$ for x in the interval [0,2pi)?

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How do you solve $sin x \cdot cos x = \frac{1}{2}$ $sinx*cosx=1/2$ for x in the interval [0,2pi)?

1 Answer

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Answer 1 · 2016-01-31T05:50:25

$x = \frac{π}{4}$ $x = pi/4$

Explanation:

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

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

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How do you solve $sin x \cdot cos x = \frac{1}{2}$ $sinx*cosx=1/2$ for x in the interval [0,2pi)?

1 Answer

Explanation:

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How do you solve $sin x \cdot cos x = \frac{1}{2}$ $sinx*cosx=1/2$ for x in the interval [0,2pi)?