Question

Another trig equation XD

Answer 1

\[2\sin(2x) - 2sinx + 2\sqrt{3}cosx - \sqrt{3} = 0\]

Answer 2

Where do I even start with this one..

Answer 3

Identity:
\[\Large \sin(2\theta) = 2\sin(\theta)\cos(\theta)\]


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Use the identity above to break up the sin(2x). Then factor by grouping


\[\Large 2\sin(2x) - 2\sin(x) + 2\sqrt{3}\cos(x)-\sqrt{3} = 0\]


\[\Large 2*2\sin(x)\cos(x) - 2\sin(x) + 2\sqrt{3}\cos(x)-\sqrt{3} = 0\]


\[\Large \left(2*2\sin(x)\cos(x) - 2\sin(x)\right) + \left(2\sqrt{3}\cos(x)-\sqrt{3}\right) = 0\]


\[\Large 2\sin(x)\left(2\cos(x) - 1\right) + \left(2\sqrt{3}\cos(x)-\sqrt{3}\right) = 0\]


\[\Large 2\sin(x)\left(2\cos(x) - 1\right) + \sqrt{3}\left(2\cos(x)-1\right) = 0\]


\[\Large \left(2\sin(x)+\sqrt{3}\right)\left(2\cos(x) - 1\right) = 0\]


I'll let you take over from here.

Answer 4

Thank you so much! I'll let you know what I get...

Answer 5

I got 4pi/3, 5pi/3, pi/3 and 5pi/3 again.
How would I write all of that?

Like this?
4pi/3 + 2npi
5pi/3 + 2npi
pi/3 + 2npi

Is there any way to condense it more?

Answer 6

I think that's as condensed as it's going to get. Nice work.

Answer 7

Thanks a bunch!

Answer 8

sure thing

Answer 9

pi/3 and 4pi/3 are pi apart from one another.

So you condense it a tad bit if you like :)

5pi/3 + 2kpi
pi/3 + kpi

Answer 10

oo, nice! thanks zep :)