SOLUTION: In this problem, we find all x with 0<x<2pi such that tanx +secx =rad3 Use sin^2x+cos^2x=1 to find cosx.

Algebra ->  Test -> SOLUTION: In this problem, we find all x with 0<x<2pi such that tanx +secx =rad3 Use sin^2x+cos^2x=1 to find cosx.      Log On


   

Question 1178808: In this problem, we find all x with 0 Use sin^2x+cos^2x=1 to find cosx.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

given: tan%28x%29+%2Bsec%28x%29+=sqrt%283%29 and 0%3C+x+%3C+2pi

We use the identity: sin%5E2%28x%29+%2Bcos%5E2%28+x+%29=+1 to find cos%28x%29

cos%5E2%28+x+%29=+1-sin%5E2%28x%29+..........factor

cos%5E2%28+x+%29=%28+1-sin%28x%29%29%28+1%2Bsin%28x%29%29

cos%5E2%28+x+%29%2F%28+1-sin%28x%29%29=%28+1%2Bsin%28x%29%29.....1)


tan%28x%29=sin%28x%29%2Fcos%28x%29 and sec%28x%29=1%2Fcos%28x%29


sin%28x%29%2Fcos%28x%29+%2B1%2Fcos%28x%29+=sqrt%283%29

%28sin%28x%29%2B1%29%2Fcos%28x%29+=sqrt%283%29.......substitute %28+1%2Bsin%28x%29%29 from 1)

%28cos%5E2%28+x+%29%2F%28+1-sin%28x%29%29%29%2Fcos%28x%29+=sqrt%283%29

cos%28x%29%2F%281+-+sin%28x%29%29=sqrt%283%29

use the identity: sin%5E2%28x%29+%2Bcos%5E2%28+x+%29=+1 to find sin%28x%29

sin%28x%29=sqrt%281-cos%5E2%28+x+%29%29

cos%28x%29%2F%281+-+sqrt%281-cos%5E2%28+x+%29%29%29=sqrt%283%29

cos%28x%29=%281+-+sqrt%281-cos%5E2%28+x+%29%29%29%2Asqrt%283%29

cos%28x%29%2Fsqrt%283%29=1+-+sqrt%281-cos%5E2%28+x+%29%29

cos%28x%29%2Fsqrt%283%29-1=+-+sqrt%281-cos%5E2%28+x+%29%29.....square both sides

%28cos%28x%29%2Fsqrt%283%29-1%29%5E2=+%28-+sqrt%281-cos%5E2%28+x+%29%29%29%5E2

%28cos%28x%29-sqrt%283%29%29%2F%28sqrt%283%29%29%5E2=+1-cos%5E2%28+x+%29

%28cos%28x%29-sqrt%283%29%29%5E2%2F3=+1-cos%5E2%28+x+%29

cos%5E2%28x%29-2cos%28x%29%2Asqrt%283%29%2B3=+3-3cos%5E2%28+x+%29

cos%5E2%28x%29%2B3cos%5E2%28+x+%29-2cos%28x%29%2Asqrt%283%29%2B3-+3=0

4cos%5E2%28+x+%29-2cos%28x%29%2Asqrt%283%29=0

2cos%28x%29%282cos%28+x+%29-sqrt%283%29%29=0

solutions:

if 2cos%28x%29=0=> cos%28x%29=0
or
if 2cos%28+x+%29-sqrt%283%29=0=>2cos%28+x+%29=sqrt%283%29=>cos%28+x+%29=sqrt%283%29%2F2


Since tan+%28x%29 and cos+%28x%29 are positive, then x lies in the first quadrant, and in sin%28x%29%2Fcos%28x%29+%2B1%2Fcos%28x%29+=sqrt%283%29

cos%28x%29=0 is denominator, so disregard this solution

cos+%28x+%29=+sqrt%283%29%2F2

x=+cos%5E-1%28sqrt%283%29%2F2+%29

x=+pi%2F6 (result in radians)-> your solution

convert to degrees

+x+=+30° (degrees)


Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Had your post be presented accurately,  I would show/teach you,  how to solve this  (and similar)  problem/problems.

But for the person who presents  Math problem in this way,  I will do  NOTHING.


----------------


Regarding the solution by @MathLover1,  I can say, that for a standard/routine/regular
school Math problem,  the solution should not be longer than  10  lines of the standard text  (as an average).


If it is significantly longer,  I'd think that something is out of the order in the solution.

I'd say then,  that something is out of the order  EITHER  in the solution itself,  OR  in its methodology,
OR  in the way of teaching.


Actually,  there is a clear,  straightforward classic elegant solution
to this and similar problems in  5 - 6  lines long.