MHB Eaglesfan1717's question at Yahoo Answers regarding a trigonometric equation

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The trigonometric equation sin(x) = 2 sin(x) cos(x) can be rearranged and factored to sin(x)(2 cos(x) - 1) = 0. This leads to two sets of solutions: sin(x) = 0, resulting in x = kπ, and 2 cos(x) - 1 = 0, giving cos(x) = 1/2, which leads to x = ±π/3 + 2kπ. The discussion encourages users to engage further with trigonometry questions on a dedicated math forum. The solutions provided are essential for understanding the roots of the given equation. This concise analysis aids in solving similar trigonometric problems.
MarkFL
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Here is the question:

Help with trig equation :)?

sin x = 2 sin x cos x

Here is a link to the question:

Help with trig equation :)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello eaglesfan1717,

We are given to solve:

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

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

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

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

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

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

$$x=k\pi$$ where $$k\in\mathbb{Z}$$.

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

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

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

To eaglesfan1717 and any other guests viewing this topic, I invite and encourage you to post your trigonometry questions in our http://www.mathhelpboards.com/f12/ forum.

Best Regards,

Mark.
 
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