# Trigonometric Equations Help!!!!!

• January 24th 2009, 07:46 PM
MP3
Trigonometric Equations Help!!!!!
Using a calculator, determine the solutions for each equation. The domain is 0≤"theta" (The 0 with the line across it) ≤ 360 degrees

a. tan x=0.75

The two possible solutions i got were 36.9 degrees and 216.9 degrees (i got 216.9 degrees by doing 180+36.9 since it was in quadrant three. BUT in my text book the answer is 323.1degrees.

Is the textbook correct or incorrect? Is my answer correct or incorrect?
If i am incorrect can you please tell me how you get 323.1degrees?

Thank you!
• January 24th 2009, 07:52 PM
Chris L T521
Quote:

Originally Posted by MP3
Using a calculator, determine the solutions for each equation. The domain is 0≤"theta" (The 0 with the line across it) ≤ 360 degrees

a. tan x=0.75

The two possible solutions i got were 36.9 degrees and 216.9 degrees (i got 216.9 degrees by doing 180+36.9 since it was in quadrant three. BUT in my text book the answer is 323.1degrees.

Is the textbook correct or incorrect? Is my answer correct or incorrect?
If i am incorrect can you please tell me how you get 323.1degrees?

Thank you!

Your solution to $\tan x=0.75$ is correct.

As a side note, if the question was $\tan x={\color{red}-}0.75$, one of the possible solutions would be 323.1 degrees [the solution given by the textbook].
• January 24th 2009, 07:55 PM
MP3
Hi, i dont quite understand. Mind if you reword or rephrase that because the question was postive 0.75. Is it suppose to be negative because im quite lost thank you

Quote:

Originally Posted by Chris L T521
For what you have given us, you're answers are correct.

However... it the problem was $\tan x={\color{red}-}0.75$ then one of the solutions would be 323.1 degrees!!

• January 25th 2009, 02:12 PM
masters
Quote:

Originally Posted by MP3
Hi, i dont quite understand. Mind if you reword or rephrase that because the question was postive 0.75. Is it suppose to be negative because im quite lost thank you

Hi MP3,

All Chris is saying is that your original calculations were correct if, in fact, the original equation was:

$\tan x=0.75$

$\tan x = -0.75$