Question

Sorry, I typed the problem wrong.... Here is the proper equation.

\[\cos \theta=-\sqrt{3}\sin\theta]

Answer 1

\[\large \cos \theta=-\sqrt{3}\sin\theta\]

Answer 2

Evidently I still can't type it properly =P

Answer 3

well, is what i wrote correct?

Answer 4

Yes, thank you.... Sorry.

Answer 5

\[\cos\theta+\sqrt{3} \sin\theta=0 \]

Answer 6

Could I have said \[\cos \theta/\sin \theta=-\sqrt{3} \]

Answer 7

Is then equal to cot=sqrt -3?

Answer 8

You can use the trig transformation:\[acos\theta+bsin\theta=\sqrt{a^{2}+b^{2}} \cos\left(\theta-\alpha\right) \]

Answer 9

However, the method you stated would work as well.

Answer 10

Could I have then said \[\tan \theta=-\sqrt{3}/3\]

Answer 11

Thank you very much for your help by the way.

Answer 12

\[-\sqrt{3} = \frac{\cos\theta}{\sin\theta} => \frac{\sin\theta}{\cos\theta} = -\frac{1}{\sqrt{3}} => \tan\theta = -\frac{1}{\sqrt{3}} => \theta= \arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]

Answer 13

Assuming its in radians.