intersections of 3sin(2x) and y=2

**Stuck Man** · February 7th 2010, 06:08 AM

I need to find the intersections of 3sin(2x) and y=2. I can expand sin(2x) with a double angle formula but I'm not getting the answers. The range is 0 - 360.

**Stuck Man** · February 7th 2010, 06:48 AM

I presume the reply is being changed. Its not what I was expecting. The book I'm using has not taught me to do questions that way.

**HallsofIvy** · February 7th 2010, 07:22 AM

Originally Posted by Stuck Man

I need to find the intersections of 3sin(2x) and y=2. I can expand sin(2x) with a double angle formula but I'm not getting the answers. The range is 0 - 360.

I can see no reason to use a double angle formula: y= 3sin(2x)= 2 so sin(2x)= 2/3. Use a calculator to find $2x= sin^{-1}(2/3)$ and then divide by 2. sin(180- x)= sin(x) so to get the other value, subtract the 2x your calculator gives you from 180.

**Stuck Man** · February 7th 2010, 07:42 AM

I don't think the last part is correct. The second intersection is at 90 - x. There are two more between 180 and 270.

**Stuck Man** · February 7th 2010, 08:09 AM

The question can be done with the double angle formula with tangent. My book hasn't covered that.

**skeeter** · February 7th 2010, 11:54 AM

Originally Posted by Stuck Man

I need to find the intersections of 3sin(2x) and y=2. I can expand sin(2x) with a double angle formula but I'm not getting the answers. The range is 0 - 360.

since $0 < x < 360$

$0 < 2x < 720$

$3\sin(2x) = 2$

$\sin(2x) = \frac{2}{3}$

$2x = \arcsin\left(\frac{2}{3}\right)$

$2x = 180 - \arcsin\left(\frac{2}{3}\right)$

$2x = 360 + \arcsin\left(\frac{2}{3}\right)$

$2x = 540 - \arcsin\left(\frac{2}{3}\right)$

solve for $x$ in all four equations

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