Solving trigonometric equations

**IDontunderstand** · January 5th 2011, 03:41 PM

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

**dwsmith** · January 5th 2011, 03:44 PM

$sec(2tan+sec)=0$

**IDontunderstand** · January 5th 2011, 03:47 PM

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

**dwsmith** · January 5th 2011, 03:50 PM

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?

**pickslides** · January 5th 2011, 03:55 PM

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

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

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

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

Can you finish it?

**IDontunderstand** · January 5th 2011, 04:03 PM

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

**Archie Meade** · January 5th 2011, 04:07 PM

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.

**IDontunderstand** · January 5th 2011, 04:14 PM

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

**IDontunderstand** · January 5th 2011, 04:16 PM

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.

**Archie Meade** · January 5th 2011, 04:17 PM

Yes

**IDontunderstand** · January 5th 2011, 04:19 PM

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.

**Archie Meade** · January 5th 2011, 04:27 PM

Originally Posted by IDontunderstand

If we say "x" is secant whih is zero, then I have to find arcsecant of zero, which is undefinded.

For your 2 solutions x,

$cosx\ \ne\ 0$

so those solutions are fine, you don't have 0/0

**ericpepin** · January 15th 2011, 01:55 AM

Great solution. Thanks

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