Question 112355
*[Tex \LARGE 2\tan^2(x)-5\sec(x)-10=0] Start with the given equation



*[Tex \LARGE 2(\sec^2(x)-1)-5\sec(x)-10=0] Transform the identity *[Tex \Large 1+\tan^2(x)=\sec^2(x)] to *[Tex \Large \tan^2(x)=\sec^2(x)-1]. Now replace *[Tex \Large \tan^2(x)] with *[Tex \Large \sec^2(x)-1]



*[Tex \LARGE 2\sec^2(x)-2-5\sec(x)-10=0] Distribute



*[Tex \LARGE 2\sec^2(x)-5\sec(x)-12=0] Combine like terms



Now let *[Tex \LARGE w=\sec(x)]. So we now get


*[Tex \LARGE 2w^2-5w-12=0]



Let's use the quadratic formula to solve for w:



Starting with the general quadratic


{{{aw^2+bw+c=0}}}


the general solution using the quadratic equation is:


{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{2*w^2-5*w-12=0}}} ( notice {{{a=2}}}, {{{b=-5}}}, and {{{c=-12}}})





{{{w = (--5 +- sqrt( (-5)^2-4*2*-12 ))/(2*2)}}} Plug in a=2, b=-5, and c=-12




{{{w = (5 +- sqrt( (-5)^2-4*2*-12 ))/(2*2)}}} Negate -5 to get 5




{{{w = (5 +- sqrt( 25-4*2*-12 ))/(2*2)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{w = (5 +- sqrt( 25+96 ))/(2*2)}}} Multiply {{{-4*-12*2}}} to get {{{96}}}




{{{w = (5 +- sqrt( 121 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{w = (5 +- 11)/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{w = (5 +- 11)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{w = (5 + 11)/4}}} or {{{w = (5 - 11)/4}}}


Lets look at the first part:


{{{x=(5 + 11)/4}}}


{{{w=16/4}}} Add the terms in the numerator

{{{w=4}}} Divide


So one answer is

{{{w=4}}}




Now lets look at the second part:


{{{x=(5 - 11)/4}}}


{{{w=-6/4}}} Subtract the terms in the numerator

{{{w=-3/2}}} Divide


So another answer is

{{{w=-3/2}}}


So our possible solutions are:

{{{w=4}}} or {{{w=-3/2}}}




Remember, we let *[Tex \LARGE w=\sec(x)]. So *[Tex \LARGE \sec(x)=4] or *[Tex \LARGE \sec(x)=-3/2] 



This means that *[Tex \LARGE \frac{1}{\cos(x)}=4]===>*[Tex \LARGE \cos(x)=1/4]



and *[Tex \LARGE \frac{1}{\cos(x)}=-3/2]===>*[Tex \LARGE \cos(x)=-2/3]



Now solve for x in each case


*[Tex \LARGE \cos(x)=1/4] Start with the first solution



*[Tex \LARGE x=arccos(1/4)] Take the arccosine of both sides



*[Tex \LARGE x=1.31811607165281] Take the arccosine of {{{1/4}}}


However, this angle (which is about 75.5 degrees) is in the first quadrant. So it does not satisfy the original conditions




*[Tex \LARGE \cos(x)=-2/3] Now move onto the second solution



*[Tex \LARGE x=arccos(-2/3)] Take the arccosine of both sides



*[Tex \LARGE x=2.30052398302187] Take the arccosine of {{{-2/3}}}


Converting to degrees, we get about 131.8 degrees, which is in the second quadrant



So the solution is 


*[Tex \LARGE x=2.30052398302187+2\pi n]