Solve forx, 0≤x≤360. Give exact values for x

tan (3x-(pie/3))=-1

Results 1 to 3 of 3

- Feb 19th 2006, 11:39 PM #1

- Joined
- Feb 2006
- Posts
- 14

- Feb 20th 2006, 02:11 AM #2

- Joined
- Apr 2005
- Posts
- 1,631

Originally Posted by**rachael**

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

- Feb 20th 2006, 10:51 AM #3

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Originally Posted by**rachael**

$\displaystyle x=-\frac{\pi}{4}+\pi k$ where $\displaystyle k$ is some integer.

Thus, $\displaystyle \tan \left(3x-\frac{\pi}{3}\right)=-1$

becomes, $\displaystyle 3x-\frac{\pi}{3}=-\frac{\pi}{4}+\pi k$.

Thus,

$\displaystyle x=\frac{\pi}{36}+\frac{\pi}{3}k$

But we need that,

$\displaystyle 0\leq x\leq 2\pi$, thus,

$\displaystyle 0\leq \frac{\pi}{36}+\frac{\pi}{3}k\leq 2\pi$

Thus, by solving the inequality.

$\displaystyle -.083\bar 3=-\frac{1}{12}\leq k\leq\frac{71}{12}= 5.91\bar 6$

Thus, since $\displaystyle k$ is integer we have that.

$\displaystyle k=0,1,2,3,4,5$.

Thus, since $\displaystyle x=\frac{\pi}{36}+\frac{\pi}{3} k=-5^o+60^ok$

We have that,

$\displaystyle x=5^o$

$\displaystyle x=65^o$

$\displaystyle x=125^o$

$\displaystyle x=185^o$

$\displaystyle x=245^o$

$\displaystyle x=305^o$

One for each value of $\displaystyle k$.

Q.E.D.

Click on a term to search for related topics.