# Trigonometric Equations

• Jan 6th 2013, 06:52 PM
maca404
Trigonometric Equations
I have been given the problem $\displaystyle 3/tanx=2sinx$ I can solve this easily graphically however I am lost algebraically, I think there is squaring and trigonometrical identities involved but this whole thing has really ground me to a halt I have no idea were to start. If someone could give me some pointers that would be great.
• Jan 6th 2013, 07:35 PM
skeeter
Re: Trigonometric Equations
Quote:

Originally Posted by maca404
I have been given the problem $\displaystyle 3/tanx=2sinx$ I can solve this easily graphically however I am lost algebraically, I think there is squaring and trigonometrical identities involved but this whole thing has really ground me to a halt I have no idea were to start. If someone could give me some pointers that would be great.

$\displaystyle \frac{3\cos{x}}{\sin{x}}=\frac{2\sin^2{x}}{\sin{x} }$ , note $\displaystyle \sin{x} \ne 0$

$\displaystyle 3\cos{x}=2(1-\cos^2{x})$

$\displaystyle 2\cos^2{x}+3\cos{x}-2=0$

$\displaystyle (2\cos{x}-1)(\cos{x}+2)=0$

Can you finish?
• Jan 6th 2013, 07:38 PM
ibdutt
Re: Trigonometric Equations
Just do little bit calculations and proceed
3 = 2 sin x tanx = 2 sinx ( sinx / cos x ) = 2 sin^2 x / cosx
So we get 3cosx = 2 sin^2 x
Now convert sin2x in terms of cos^2 x and you have a quadratic in cos x.
I am sure you can now solve it further.
• Jan 6th 2013, 10:11 PM
maca404
Re: Trigonometric Equations
Ok I see what everyone is getting at now, I will put what I have below but I need to work out the equation from 0 to 2pi. I have the first one worked out not sure how to get the second:

$\displaystyle (2cosx-1)(cosx+2)=0$

$\displaystyle cosx + 2 = 0$

$\displaystyle cosx = -2$

$\displaystyle No Solution$

$\displaystyle 2 cosx -1 =0$

$\displaystyle 2 cosx = 1$

$\displaystyle cosx = 1/2$

$\displaystyle x = cos^-1(1/2)$

$\displaystyle x = 1.04 rad$

Now I know there is another but I am usure how to find it ?
• Jan 6th 2013, 11:15 PM
grillage
Re: Trigonometric Equations
If you are working up to $\displaystyle 2\pi$ then there will be another angle in the circle that gives $\displaystyle cosx=\frac{1}{2}$. Your calculator will only give you the one from 0 to $\displaystyle \pi$
• Jan 6th 2013, 11:47 PM
maca404
Re: Trigonometric Equations
I can see graphically that I can find the next angle by simply doing 2pi - 1.04 , Is this is the corret way to do it ?
• Jan 7th 2013, 12:13 AM
grillage
Re: Trigonometric Equations
Yes, that sounds good to me.
• Jan 7th 2013, 02:50 AM
skeeter
Re: Trigonometric Equations
$\displaystyle \cos{x}=\frac{1}{2}$ at $\displaystyle x=\frac{\pi}{3}$ and $\displaystyle x=\frac{5\pi}{3}$