# 3cosx+4tanx+secx=0

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• Feb 24th 2011, 11:08 AM
jonnygill
3cosx+4tanx+secx=0
hello,

i'm trying to work out this problem to solve for Theta...

$\displaystyle 3cos\Theta+4tan\Theta+sec\Theta=0$

I decided it would make sense to get all trig functions to be in terms of sec.

So after manipulating the equation i came up with...

$\displaystyle 15sec^4\Theta-22sec^2\Theta=9$

or

$\displaystyle 15sec^4\Theta-22sec^2\Theta-9=0$

i tried manipulating it a bit more and got...

$\displaystyle 9cos^4\Theta+22cos^2\Theta-15=0$

i kind of keep going on circles with this one. is there something i'm missing? perhaps i approached this problem the wrong way?

thanks!
• Feb 24th 2011, 11:19 AM
harish21
use $\displaystyle \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta}$

and $\displaystyle \sec(\theta)=\dfrac{1}{\cos(\theta)}$

and simplify further
• Feb 24th 2011, 11:27 AM
jonnygill
Quote:

Originally Posted by harish21
use $\displaystyle \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta}$

and $\displaystyle \sec(\theta)=\dfrac{1}{\cos(\theta)}$

and simplify further

ok, just did that and got...

$\displaystyle 3cos\Theta+\dfrac{4}{cos^2\Theta}+\dfrac{1}{cos\Th eta}=0$

i'm not sure how to proceed from here
• Feb 24th 2011, 11:33 AM
topsquark
Quote:

Originally Posted by jonnygill
$\displaystyle 9cos^4\Theta+22cos^2\Theta-15=0$

Let x = cos^2(theta). Then your equation is 9x^2 + 22x - 15 = 0. Solve for x. Then use cos(theta) = sqrt(x).

-Dan
• Feb 24th 2011, 11:59 AM
jonnygill
Quote:

Originally Posted by topsquark
Let x = cos^2(theta). Then your equation is 9x^2 + 22x - 15 = 0. Solve for x. Then use cos(theta) = sqrt(x).

-Dan

i made this substitution, and used the quadratic formula to find possible values of x. i ignored the solution where x=-6, since we will be taking the sqrt of a negative number, and i'm not interested in non-real solutions for this problem. so, i got $\displaystyle cos\Theta=\pm\sqrt\dfrac{5}{9}$.

i solved for Theta and got... $\displaystyle \Theta=138.1897$ OR $\displaystyle \Theta=41.8103$

i plugged both of these into the original equation and neither one worked!!

what did i do wrong???
• Feb 24th 2011, 12:49 PM
Archie Meade
Quote:

Originally Posted by jonnygill
hello,

i'm trying to work out this problem to solve for Theta...

$\displaystyle 3cos\Theta+4tan\Theta+sec\Theta=0$

I decided it would make sense to get all trig functions to be in terms of sec.

So after manipulating the equation i came up with...

$\displaystyle 15sec^4\Theta-22sec^2\Theta=9$

or

$\displaystyle 15sec^4\Theta-22sec^2\Theta-9=0$

i tried manipulating it a bit more and got...

$\displaystyle 9cos^4\Theta+22cos^2\Theta-15=0$

i kind of keep going on circles with this one. is there something i'm missing? perhaps i approached this problem the wrong way?

thanks!

Crikey jonny!

you really seem to like making things hard for yourself!

Here's another way you could do it.
Express in terms of simplest trig functions, sine and cosine..

$\displaystyle \displaystyle\ 3cos\Theta+4\frac{sin\Theta}{cos\Theta}+\frac{1}{c os\Theta}=0$

obtain a common denominator, after which the numerator must be zero...

$\displaystyle \displaystyle\frac{3cos^2\Theta}{cos\Theta}+4\frac {sin\Theta}{cos\Theta}+\frac{1}{cos\Theta}=0$

$\displaystyle \displaystyle\frac{3cos^2\Theta+4sin\Theta+1}{cos\ Theta}=0$

$\displaystyle 3\left(1-sin^2\Theta\right)+4sin\Theta+1=0$

$\displaystyle 3-3sin^2\Theta}+4sin\Theta+1=0$

$\displaystyle 3sin^2\Theta-4sin\Theta-4=0$

This factors into

$\displaystyle \left(3sin\Theta+2)(sin\Theta-2)=0$

and one of those two factors can never be zero.
• Feb 24th 2011, 01:29 PM
topsquark
Quote:

Originally Posted by jonnygill
i made this substitution, and used the quadratic formula to find possible values of x. i ignored the solution where x=-6, since we will be taking the sqrt of a negative number, and i'm not interested in non-real solutions for this problem. so, i got $\displaystyle cos\Theta=\pm\sqrt\dfrac{5}{9}$.

i solved for Theta and got... $\displaystyle \Theta=138.1897$ OR $\displaystyle \Theta=41.8103$

i plugged both of these into the original equation and neither one worked!!

what did i do wrong???

The reference angle is about 41 degrees. So far so good. Now, put that reference angle in the 4th quadrant and what do you get?

-Dan