1. ## 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

2. 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
$tan(2x - 45) = 1$

The inverse tangent of 1 is 45 or 225 degrees.

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

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

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

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

3. Thankyou!!!!!!!!!!!!!!!! Thats great.

4. Sorry to be annoying, but how do you get 225degrees

5. Originally Posted by orangesmarties
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

6. Thanks again Dan- thats been a great help!!

7. Hello, orangesmarties!

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

Answers: . $x \;= \;45^o,\:135^o,\:225^o,\:315^o$
We have: . $\tan(2x-45^o) \:=\:1$

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

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

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

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

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

8. Thanks Soroban!