How do you solve $sin 2 x cos x - cos 2 x sin x = \frac{1}{2}$ $Sin2x cosx - cos2x sinx = 1/2$ ?

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How do you solve $sin 2 x cos x - cos 2 x sin x = \frac{1}{2}$ $Sin2x cosx - cos2x sinx = 1/2$ ?

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Answer 1 · 2016-02-21T06:02:25

$x = \frac{π}{6}, \frac{5 π}{6}$ $x=pi/6, (5pi)/6$ or

$x = 30^{\circ}, 150^{\circ}$ $x=30^@, 150^@$

Explanation:

From sum and difference formulas

We have

$sin (x + y) = sin x cos y + cos x sin y$ $sin (x+y)=sin x cos y + cos x sin y$
$sin (x - y) = sin x cos y - cos x sin y$ $sin (x-y)=sin x cos y - cos x sin y$

$sin (2 x - x) = sin 2 x cos x - cos 2 x sin x$ $sin (2x-x)=sin 2x cos x - cos 2x sin x$

$sin x = sin 2 x cos x - cos 2 x sin x$ $sin x=sin 2x cos x - cos 2x sin x$

So

$sin x = \frac{1}{2}$ $sin x=1/2$

$x = {sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}, \frac{5 π}{6}$ $x=sin^-1 (1/2)=pi/6, (5pi)/6$

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How do you solve $sin 2 x cos x - cos 2 x sin x = \frac{1}{2}$ $Sin2x cosx - cos2x sinx = 1/2$ ?

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Explanation:

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How do you solve Sin2x cosx - cos2x sinx = 1/2sin2xcosx−cos2xsinx=12Sin2x cosx - cos2x sinx = 1/2?

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Explanation:

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How do you solve $sin 2 x cos x - cos 2 x sin x = \frac{1}{2}$ $Sin2x cosx - cos2x sinx = 1/2$ ?