0
$\begingroup$

How do you solve this trigonometric equation?

Solve the equation for solutions in the interval $[0,2 \pi)$. $$\left(\cot(x) -1 \right) \left( 2 \sin(x) + \sqrt{3} \right) = 0.$$

$\endgroup$
3
  • 1
    $\begingroup$ Hint: if $a \cdot b = 0$, then $a = 0$ or $b = 0$. Solve both equations to get all solutions. $\endgroup$ May 6, 2012 at 23:35
  • 2
    $\begingroup$ Any particular reason you need to use offensive language in your user name? $\endgroup$ May 6, 2012 at 23:38
  • 1
    $\begingroup$ The profanity in the user's display name has been removed. @I'm just some guy: Any future use of profanity will be cause for suspension. $\endgroup$ May 6, 2012 at 23:50

1 Answer 1

2
$\begingroup$

A product of two real valued quantities is zero if and only if one of the terms in the product is zero.

So, here, either $$ \cot x-1=0 $$ or $$ 2\sin x +\sqrt 3=0. $$ Equivalently either $\cot x=1$ or $\sin x =-\sqrt 3/2$. Can you find the solutions to these? Remember to only take solutions in $[0,2\pi)$.

$\endgroup$
1
  • $\begingroup$ Yep, I found the correct solution set as {pi/4, 5pi/4, 4pi/3, 5pi/3}. Thanks! $\endgroup$
    – Curious
    May 6, 2012 at 23:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .