# tan equation

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• May 13th 2008, 10:14 AM
orangesmarties
tan equation
Solve for 0<x<360, the equation tan(2x - 45) = 1

(N.B. the question is in degrees & the interval is greater than or EQUAL to 0 and less than OR EQUAL to 360)

The solution is:
x = 45, 135, 225, 315

I am just not sure how to get to the answer-help would really be appreciated- BIG exam in less than a day. Thanks x
• May 13th 2008, 11:12 AM
topsquark
Quote:

Originally Posted by orangesmarties
Solve for 0<x<360, the equation tan(2x - 45) = 1

(N.B. the question is in degrees & the interval is greater than or EQUAL to 0 and less than OR EQUAL to 360)

The solution is:
x = 45, 135, 225, 315

I am just not sure how to get to the answer-help would really be appreciated- BIG exam in less than a day. Thanks x

$\displaystyle tan(2x - 45) = 1$

The inverse tangent of 1 is 45 or 225 degrees.

Thus
$\displaystyle 2x - 45 = 45 \implies x = 45$
and
$\displaystyle 2x - 45 = 225 \implies x = 135$

To get the other two solutions note that
$\displaystyle tan(45 + 360) = 1$
and
$\displaystyle tan(225 + 360) = 1$

Thus
$\displaystyle 2x - 45 = 405 \implies x = 225$
and
$\displaystyle 2x - 45 = 585 \implies x = 315$

We would do this again, eg tan(45 + 2 * 360) = 1, but the solutions for x come out to be greater than 360, so we are done.

-Dan
• May 13th 2008, 11:15 AM
orangesmarties
Thankyou!!!!!!!!!!!!!!!! Thats great.
• May 13th 2008, 11:22 AM
orangesmarties
(Worried)Sorry to be annoying, but how do you get 225degrees
• May 13th 2008, 11:24 AM
topsquark
Quote:

Originally Posted by orangesmarties
(Worried)Sorry to be annoying, but how do you get 225degrees

The reference angle for tan(x) = 1 is x = 45. Tangent is positive in QI and QIII. So the possible values for x between 0 and 360 are 45 and 180 + 45 = 225.

-Dan
• May 13th 2008, 11:33 AM
orangesmarties
Thanks again Dan- thats been a great help!!
• May 13th 2008, 11:36 AM
Soroban
Hello, orangesmarties!

Quote:

Solve: .$\displaystyle \tan(2x-45^o) \:=\;1,\quad 0 \leq x \leq 360^o$

Answers: .$\displaystyle x \;= \;45^o,\:135^o,\:225^o,\:315^o$

We have: .$\displaystyle \tan(2x-45^o) \:=\:1$

Then: .$\displaystyle 2x-45^o \;=\;45^o + 180^on$ . for some integer $\displaystyle n.$

And: .$\displaystyle 2x \;=\;90^o + 180^on$

. . Hence: .$\displaystyle x \;=\;45^o + 90^on$

Let $\displaystyle n \:=\:0,1,2,3$

. . and we have: .$\displaystyle \boxed{x \;=\;45^o,\:135^o,\:225^o,\:315^o}$

• May 17th 2008, 02:28 AM
orangesmarties
Thanks Soroban!