SOLUTION: How do you solve over an interval of [0, 2π)
{{{ (cotx-1)(sqrt(3)cotx+1) = 0 }}}
So far with my work I got,
{{{ cotx-1=0 }}} and {{{ (sqrt(3)cotx+1) = 0 }}} to solve fo
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-> SOLUTION: How do you solve over an interval of [0, 2π)
{{{ (cotx-1)(sqrt(3)cotx+1) = 0 }}}
So far with my work I got,
{{{ cotx-1=0 }}} and {{{ (sqrt(3)cotx+1) = 0 }}} to solve fo
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Question 1114976: How do you solve over an interval of [0, 2π)
So far with my work I got, and to solve for
cotx = 1
cosx/sinx = 1
cosx = 1
this would = π , as a solution
cotx = -1/sqrt(3)
I get lost on how to handle to second part or
I would appreciate some help on this! Thank you! Answer by greenestamps(13200) (Show Source):
You are making these harder than they are. All the solutions are the common points on the unit circle.
Your work on the first one...
cotx-1=0
cotx = 1 yes...
cosx/sinx = 1 yes; but not needed
cosx = 1 no; this doesn't follow from your previous equation
this would = pi; as a solution no; cosx=1 at 0, not at pi
After
cotx=1
you could simply do
tanx=1
after which you should know
x = pi/4 or x = 5pi/4
Your work on the second one...
(sqrt(3)cotx+1) = 0
cotx = -1/sqrt(3)
From there you should go to
tanx = -sqrt(3)
and from there you should know
x = 2pi/3 or x = 4pi/3
