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eaglesfan1717's question at Yahoo! Answers regarding a trigonometric equation

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello eaglesfan1717,

We are given to solve:

\(\displaystyle \sin(x)=2\sin(x)\cos(x)\)

I would arrange the equation so that we may factor and utilize the zero-factor property:

\(\displaystyle 2\sin(x)\cos(x)-\sin(x)=0\)

\(\displaystyle \sin(x)(2\cos(x)-1)=0\)

Equating the factors in turn to zero yields the following roots:

i) \(\displaystyle \sin(x)=0\)

\(\displaystyle x=k\pi\) where \(\displaystyle k\in\mathbb{Z}\).

ii) \(\displaystyle 2\cos(x)-1=0\)

\(\displaystyle \cos(x)=\frac{1}{2}\)

\(\displaystyle x=\pm\frac{\pi}{3}+2k\pi=\frac{\pi}{3}(6k\pm1)\)

To eaglesfan1717 and any other guests viewing this topic, I invite and encourage you to post your trigonometry questions in our Trigonometry forum.

Best Regards,

Mark.