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How do you solve $sin 3 x = cos 3 x \cdot tan 3 x$ $sin3x=cos3x*tan3x$ for x in the interval [0,2pi)?

1 Answer

Apr 26, 2016

The given equation is true for all values in the range $[0, 2 π)$ $[0,2pi)$

Explanation:

You could (temporarily) replace $3 x$ $3x$ with $θ$ $theta$
so the given equation would look like:
$XXX sin (θ) = cos (θ) \cdot tan (θ)$ $color(white)("XXX")sin(theta)=cos(theta)*tan(theta)$

But we know that $tan (θ) = \frac{sin (θ)}{cos (θ)}$ $tan(theta)=(sin(theta))/(cos(theta))$

So we have that the given equation is equivalent to
$color(white)("XXX")sin(theta)=(cancel(cos(theta)))/color(white)("X")*(sin(theta))/(cancel(cos(theta)))$

Which is true for all values of $theta$ and therefore true for all values of $x$ .

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