Thread: Tri

**rachael** · February 19th 2006, 11:39 PM

Solve for x , 0≤x≤360. Give exact values for x
tan (3x-(pie/3))=-1

**ticbol** · February 20th 2006, 02:11 AM

Originally Posted by rachael

Solve for x , 0≤x≤360. Give exact values for x
tan (3x-(pie/3))=-1

The interval for the answer is in degrees, while the pi/3 is in radians. Umm.

Here is one way.

tan[3x -pi/3] = -1

Use the trig identity
tan(A-B) = (tanA -tanB)/(1 +tanA*tanB)

So,
tan[3x -pi/3] = -1
[tan(3x) -tan(pi/3)]/[1 +tan(3x)*tan(pi/3)] = -1
tan(3x) -tan(pi/3) = -1*[1 +tan(3x)*tan(pi/3)] ------------(i)

tan(pi/3) = tan(60 degrees) = sqrt(3)/1 = sqrt(3), so, to continue from (i),

tan(3x) -sqrt(3) = -1 -tan(3x)*sqrt(3)
Isolate the tan(3x)-terms,
tan(3x) +sqrt(3)*tan(3x) = -1 +sqrt(3)
Factor out the tan(3x) in the LHS,
tan(3x)*[1 +sqrt(3)] = -1 +sqrt(3)
tan(3x) = [-1 +sqrt(3)]/[1 +sqrt(3)]
Rearranging the RHS,
tan(3x) = [sqrt(3) -1]/[sqrt(3) +1]
Rationalize the denominator of the RHS,
Multiply both numerator and denominator of the RHS by the conjugate of the denominator, which is [sqrt(3) -1],

Here, we use (a-b)^2 = a^2 -2ab +b^2, and (a+b)(a-b) = a^2 -b^2,

tan(3x) = [3 -2sqrt(3) +1]/[3 -1]
tan(3x) = [4 -2sqrt(3)]/[2]
tan(3x) = 2 -sqrt(3)
tan(3x) = 0.267949192

A positive tangent value. That means angle 3x is in the 1st or 3rd quadrant.

3x = arctan(0.267949192)
3x = 15 degrees

That also means that
3x = 15deg ----in the 1st quadrant, or,
3x = 180+15 = 195deg ----in the 3rd quadrant.

Therefore,
x = 15/3 = 5 deg ----------------answer.
or, x = 195/3 = 65 deg ----------answer.

----------------------------------------------------
Check.

When x = 5deg,
tan(3x -pi/3) = -1
tan(3*5deg -60deg) =? -1
tan(15deg -60deg) =? -1
tan(-45deg) =? -1
-1 =? -1
Yes, so, OK.

When x = 65deg,
tan(3x -pi/3) = -1
tan(3*65deg -60deg) =? -1
tan(195deg -60deg) =? -1
tan(135deg) =? -1
-1 =? -1
Yes, so, OK

-----------------------------
Ooppss, I forgot that x is within the interval [0,360 degrees]

Since what was found so far were 2 acute angles only---5 and 65 degrees---, there are more values for x in the given interval or bounds or domain of x.

Since the "angle" in the question is 3x, and one revolution is 360 deg, then 360/3 = 120deg. So, keep on adding 120deg to the 5deg and 65deg for more x-values until the interval is exhausted.
Thus,
x = 5, 65, 125, 185, 245, and 305 degrees ----answers, all five of them.

Check the x=305deg, at least,....
When x = 305deg,
tan(3x -pi/3) = -1
tan(3*305deg -60deg) =? -1
tan(915deg -60deg) =? -1
tan(855deg) =? -1
-1 =? -1
Yes, so, OK

**ThePerfectHacker** · February 20th 2006, 10:51 AM

Originally Posted by rachael

Solve for x , 0≤x≤360. Give exact values for x
tan (3x-(pie/3))=-1

If $\tan x=-1$ then all solutions must be,
$x=-\frac{\pi}{4}+\pi k$ where $k$ is some integer.

Thus, $\tan \left(3x-\frac{\pi}{3}\right)=-1$
becomes, $3x-\frac{\pi}{3}=-\frac{\pi}{4}+\pi k$ .
Thus,
$x=\frac{\pi}{36}+\frac{\pi}{3}k$
But we need that,
$0\leq x\leq 2\pi$ , thus,
$0\leq \frac{\pi}{36}+\frac{\pi}{3}k\leq 2\pi$
Thus, by solving the inequality.
$-.083\bar 3=-\frac{1}{12}\leq k\leq\frac{71}{12}= 5.91\bar 6$
Thus, since $k$ is integer we have that.
$k=0,1,2,3,4,5$ .
Thus, since $x=\frac{\pi}{36}+\frac{\pi}{3} k=-5^o+60^ok$
We have that,
$x=5^o$
$x=65^o$
$x=125^o$
$x=185^o$
$x=245^o$
$x=305^o$
One for each value of $k$ .
Q.E.D.

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