1. ## [SOLVED] Trignometry and Radians

Here's the question

I have solved the first part and got the following equation, which is correct:

$\displaystyle y=(6sin2x)+3$

Now for part (ii) I have solved till

$\displaystyle sin2x=\frac{-1}{2}$

And don't know how to get the correct answer:

Spoiler:

2. Originally Posted by unstopabl3
Here's the question

I have solved the first part and got the following equation, which is correct:

$\displaystyle y=(6sin2x)+3$

Now for part (ii) I have solved till

$\displaystyle sin2x=\frac{-1}{2}$

And don't know how to get the correct answer:

Spoiler:

Hi unstopabl3,

$\displaystyle sin(2x)=-\frac{1}{2}$

the first time this occurs is in the 3rd quadrant,
as sin(angle) points out the y co-ordinate of a point on the circle.

Hence the angle 2x is

$\displaystyle {\pi}+sin^{-1}\left(\frac{1}{2}\right)$

$\displaystyle ={\pi}+\frac{{\pi}}{6}=\frac{7{\pi}}{6}$

x is half that.

3. Yea, I figured it out myself as well! Haha

Thanks for the quick and easy to understand reply!