Trig Identities

**adam182** · November 12th 2006, 04:40 PM

Solve for $0\leq x \leq 2\pi$

well i am bad at the math tool, but its 0<=x<=2pi

Cos2x = 1 - sinx

What would be the best way to set this up?

**Soltras** · November 12th 2006, 05:48 PM

Originally Posted by adam182

Solve for $0\leq x \leq 2\pi$

well i am bad at the math tool, but its 0<=x<=2pi

Cos2x = 1 - sinx

What would be the best way to set this up?

Hi Adam.

Use the identity that $\cos(2x) = 1 - 2\sin^2x$ .

Get:

$1-2\sin^2x = 1 - \sin x$ .

So $2 \sin^2 x - \sin x = 0$ .

Factor:
$\sin x \cdot (2 \sin x - 1) = 0$ .

Can you finish it from here?

**adam182** · November 12th 2006, 07:50 PM

Yes, I got to x=30 which I converted to radians...so $\pi/6$

**Soltras** · November 12th 2006, 07:59 PM

Yes $\pi/6$ is a solution.
There are several answers in that interval given $(0 \le x \le 2\pi)$ .

Note that $5\pi/6$ is also an answer, as are $0, \pi$ , and $2\pi$ .

Do you see why?

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