SOLUTION: In this problem, we find all x with 0<x<2pi such that tanx +secx =rad3 Find all x with 0<_ x < 2pi that satisfy the original equation.

Algebra ->  Test -> SOLUTION: In this problem, we find all x with 0<x<2pi such that tanx +secx =rad3 Find all x with 0<_ x < 2pi that satisfy the original equation.       Log On


   



Question 1178809: In this problem, we find all x with 0 Find all x with 0<_ x < 2pi that satisfy the original equation.
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: sec%5E2%28x%29+-+tan%5E2%28+x+%29=+1.

Factorize sec%5E2%28+x%29+-+tan%5E2%28+x%29:

%28sec+%28x+%29%2B+tan+%28x%29%29+%28sec%28+x%29+-+tan%28+x%29%29+=+1

Substitute sec%28+x%29+%2B+tan+%28x%29+=+sqrt%283%29:

sqrt%283%29+%28sec+%28x%29+-+tan+%28x%29%29+=+1

Multiply both sides by sqrt%283%29%2F3:

sec+%28x+%29-+tan+%28x%29+=+sqrt%283%29%2F3

Now we have two equations:

sec%28+x%29+%2B+tan+%28x%29+=+sqrt%283%29 ...(1)
sec+%28x%29+-+tan%28+x%29+=+sqrt%283%29%2F3 ...(2)
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add (1) and (2) , and we have:
2sec%28x+%29=+%284sqrt%283%29%29%2F3
sec%28+x%29+=+2sqrt%283%29%2F3

use identity+cos+%28x%29+=+1%2Fsec%28+x%29+
cos+%28x%29+=+1%2Fsec%28+x%29+=+3%2F%282sqrt%283%29%29+=+sqrt%283%29%2F2

subtract (1) and (2) , and we have:
2tan+%28x%29+=+2sqrt%283%29%2F3
tan+%28x%29+=+sqrt%283%29%2F3

Since tan+%28x%29 and cos+%28x%29 are positive, then x lies in the first quadrant.

Note that

cos+%28x+%29=+sqrt%283%29%2F2
x=+cos%5E-1%28sqrt%283%29%2F2+%29
x=+pi%2F6%2B360%2An
since given that 0%3C+x+%3C+2pi, solution is
x=+pi%2F6(result in radians)
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.