Question 814828: Find the exact values of the solutions of the equation that are in the interval [0,2pi)
2tan(t)-sec^2(t) = 0
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
sec and tan are connected by the identity:  . So we can get the equation in terms of tan(x) if we substitute for  :
(Note the parentheses! It is important to use them when substituting in an expression of multiple terms.) Simplifying we get:
This is an equation in quadratic form. Rearranging the terms to get it in standard form:
Since I prefer leading coefficients that are positive, I'll multiply both sides by -1:
These quadratic form equations can be difficult when you are first learning about them. It can be helpful to use a temporary variable:
Let q = tan(t)
Then 
Substituting these in we get:
This is obviously a quadratic equation. And it will factor:
(q-1)(q-1) = 0
From the Zero Product Property:
q - 1 = 0
So q = 1.
But of course we are not interested in a solution for q. We want a solution for t. So it is time to replace our temporary variable:
tan(t) = 1
And solve for t. We should recognize a tan of 1 as a special angle value. The reference angle is  . And since the 1 is positive and since tan is positive in the first and third quadrants we get a general solution of:
 (for the first quadrant)
 (for the third quadrant)
Now we find the specific solutions in the given interval. For this we try various integers for the n's in these equations until we have found all of the angles that are between 0 and  . (Note: The bracket, [, next to zero means that zero itself is included in the interval while the parenthesis, }, next to means that itself is not included.
From ...
when n = 0, 
when n = 1 (or any higher integer), t is too large for the interval.
when n is any negative integer, t is too small for the interval.
From ...
when n = 0, 
when n = 1 (or any higher integer), t is too large for the interval.
when n is any negative integer, t is too small for the interval.
So the only solutions for t that are in the given interval are and
P.S. After you have done several of these you will get more comfortable with them. Eventually you will not need a temporary variable. You will start seeing how to go directly from:
to
etc.
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