1. Trig Questions

I tried doing these and i got to imaginary numbers when i was trying to solve for x, can anyone help?

Sec^2(x)+tan(x)=3

Sin^2(x)+3Cos^2(x)=0

2. Originally Posted by imthatgirl
I tried doing these and i got to imaginary numbers when i was trying to solve for x, can anyone help?

Sec^2(x)+tan(x)=3

Sin^2(x)+3Cos^2(x)=0
sec^2(x) + tan(x) = 3 => tan^2(x) + 1 + tan(x) = 3 => tan^2(x) + tan(x) - 2 = 0 => [tan(x) + 2][tan(x) - 1] = 0 => plenty of real solutions for x.

sin^2(x) + 3cos^2(x) = 0: Restriciting to real numbers, the sum of two squares is always greater than or equal to 0. But sin^2(x) and cos^2(x) cannot simultaneously equal 0 for any real value of x. Therefore .....

3. Originally Posted by imthatgirl
I tried doing these and i got to imaginary numbers when i was trying to solve for x, can anyone help?

Sec^2(x)+tan(x)=3

Sin^2(x)+3Cos^2(x)=0

For the second case, you could do

sin^2(x) + 3cos^2(x) = 0
sin^2(x) + cos^2(x) + 2cos^2(x) = 0
2cos^2(x) = -1
2(1 - sin^2(x)) = -1
1 - sin^2(x) = -1/2
sin^2(x) = 3/2
... I figured you can go from here...

4. Originally Posted by imthatgirl
I tried doing these and i got to imaginary numbers when i was trying to solve for x, can anyone help?

Sec^2(x)+tan(x)=3
$\displaystyle \sec^2 x = \tan^2 x+1$.
Thus, $\displaystyle \tan^2 x + \tan x + 1 = 3$. This is a quadradic which you can solve.

This is Mine 83th Post!!!

5. Originally Posted by mahurshi
For the second case, you could do

sin^2(x) + 3cos^2(x) = 0
sin^2(x) + cos^2(x) + 2cos^2(x) = 0
2cos^2(x) = -1 Mr F muses: Or you could just stop here since the square of a real number can never be negative ......
2(1 - sin^2(x)) = -1
1 - sin^2(x) = -1/2
sin^2(x) = 3/2
... I figured you can go from here...
..

6. Originally Posted by mr fantastic
..
I assumed arcsin(sqrt(3/2)) would suffice for the answer.

7. Originally Posted by mahurshi
I assumed arcsin(sqrt(3/2)) would suffice for the answer.
True. But the extra work is not necessary. And in my experience when students continue with additional unnecessary working they are prone to making errors they would otherwise not make .... Those errors in turn lead to wrong answers.

This would not happen if they stopped after completing the necessary working.

For example:

Originally Posted by mahurshi
For the second case, you could do
[snip]
sin^2(x) = 3/2
... I figured you can go from here...
It would not surprise me if a student then said $\displaystyle \sin(x) = \pm \frac{\sqrt{3}}{2}$ (NB: THIS IS WRONG) or even just $\displaystyle \sin(x) = \frac{\sqrt{3}}{2}$ (NB: THIS IS WRONG) and then proceded to get real solutions .....

All comment welcome.

8. Originally Posted by mr fantastic
True. But the extra work is not necessary. And in my experience when students continue with additional unnecessary working they are prone to making errors they would otherwise not make .... Those errors in turn lead to wrong answers.

This would not happen if they stopped after completing the necessary working.

For example:

It would not surprise me if a student then said $\displaystyle \sin(x) = \pm \frac{\sqrt{3}}{2}$ (NB: THIS IS WRONG) or even just $\displaystyle \sin(x) = \frac{\sqrt{3}}{2}$ (NB: THIS IS WRONG) and then proceded to get real solutions .....

All comment welcome.

I understand. But teachers/professors typically love it when you can do that extra step and show them that you also know how to apply inverse trig functions and actually say x = so and so (since we're asked to solve for x).