1. ## Solving trigonometric equations

2Sec(x)*tan(x)+sec(x)^2=0

I know that the answer is 210 degrees from using my graph. How would I do this by hand?

I tried this.

sec(x)(2tan(x)+sec(x))=0

I get stuck here I have also changed the sec(x)^2 to Tan(x)^2+1. But then I get stuck because I still have two different trigonometric functions. Please help. Thanks

2. $sec(2tan+sec)=0$

3. what does that mean? is that the one I want to use? how do i solve the the 2tan+sec if they are different?

4. Originally Posted by IDontunderstand
what does that mean? is that the one I want to use? how do i solve the the 2tan+sec if they are different?
How would you solve x(2y+x)=0?

5. $\displaystyle 2\sec(x)\tan(x)+\sec^2(x)=0$

$\displaystyle \frac{2}{\cos(x)}\frac{\sin(x)}{\cos(x)}+\frac{1}{ \cos^2(x)}=0$

$\displaystyle \frac{2\sin(x)}{\cos^2(x)}+\frac{1}{\cos^2(x)}=0$

$\displaystyle \frac{2\sin(x)+1}{\cos^2(x)}=0$

Can you finish it?

6. I don't know I have two variables and one equation.

7. Yes, but if the left-hand side is zero, what is the numerator?
Double-check your denominator will also not be zero for the solution x.

8. Originally Posted by Archie Meade
Yes, but if the left-hand side is zero, what is the numerator?
Double-check your denominator will also not be zero for the solution x.
zero so x can be zero and y can thus be hafl the opposite of x

9. yes multiple each side by cos^2 so that I get rid of that. Then solve sinx to be -1/2 and get the angle of 210 and 330.

10. Yes

11. Originally Posted by Archie Meade
In "pickslides" post, the numerator can is zero.
If we say "x" is secant whih is zero, then I have to find arcsecant of zero, which is undefinded.

12. Originally Posted by IDontunderstand
If we say "x" is secant whih is zero, then I have to find arcsecant of zero, which is undefinded.
$cosx\ \ne\ 0$