sinxcos30-cosxsin30=sq.root(3)/2

[Robinogrl25]

sinxcos30-cosxsin30=sq.root(3)/2

Thanks!

 

[stapel]

At a guess, did the instructions perhaps say to "solve the equation"...? If so, did the instructions specify any particular interval, or are you supposed to find the "general" solution?

When you reply, please include a clear listing of everything you have tried so far, starting with the identity you applied. Thank you.

Eliz.

 

[Robinogrl25]

I'm supposed to solve for x when x is between 0 and 360.
So far I've tried making it

sin(x-30)=sq.root(3)/2

and

sq.root(3)/2 * sin(x) - 1/2 * cos(x) = sq.root(3)/2

but I forgot where to go from there.

 

[tkhunny]

sin(x-30)=sq.root(3)/2

Why are you changing from this convenient form?

\(\displaystyle sin(y) = \frac{\sqrt{3}}{2}\) at \(\displaystyle y=\frac{\pi}{3} + 2k\pi\) and \(\displaystyle y=\frac{2\pi}{3} + 2k\pi\) for k an Integer.

Now figure out the 30º and the requested Domain.

 

[skeeter]

\(\displaystyle \L \sin{(x-30)} = \frac{\sqrt{3}}{2}\)

since \(\displaystyle \L \sin{(60)} = \frac{\sqrt{3}}{2}\) and \(\displaystyle \L \sin{(120)} = \frac{\sqrt{3}}{2}\) ...

\(\displaystyle \L x-30 = 60\)

\(\displaystyle \L x-30 = 120\)

finish.

 

[Robinogrl25]

Thanks!