SOLVED Trigonometric equation?

Feb 2010
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0
If 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?

I started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?

Thanks in advance.
 

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MHF Hall of Honor
Mar 2010
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Chicago
If 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?

I started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?

Thanks in advance.
You can use identity \(\displaystyle \displaystyle \cos(\theta + \pi) = -\cos(\theta)\), to start out.

Edit: Well I suppose it should be \(\displaystyle \displaystyle \cos(\theta - \pi) = -\cos(\theta)\), same idea.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
You should know

\(\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}\)

and

\(\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}\).


So in your case

\(\displaystyle 3\sin{\left(x + \frac{\pi}{3}\right)} = 2\cos{\left(x - \frac{2\pi}{3}\right)}\)

\(\displaystyle 3\left(\sin{x}\cos{\frac{\pi}{3}} + \cos{x}\sin{\frac{\pi}{3}}\right) = 2\left(\cos{x}\cos{\frac{2\pi}{3}} + \sin{x}\sin{\frac{2\pi}{3}}\right)\)

\(\displaystyle 3\left(\frac{1}{2}\sin{x} + \frac{\sqrt{3}}{2}\cos{x}\right) = 2\left(-\frac{1}{2}\cos{x} + \frac{\sqrt{3}}{2}\sin{x}\right)\)

\(\displaystyle \frac{3}{2}\sin{x} + \frac{3\sqrt{3}}{2}\cos{x} = -\cos{x} + \sqrt{3}\sin{x}\)

\(\displaystyle \left(\frac{3}{2} - \sqrt{3}\right)\sin{x} = \left(-1 - \frac{3\sqrt{3}}{2}\right)\cos{x}\)

\(\displaystyle \left(\frac{3 - 2\sqrt{3}}{2}\right)\sin{x} = -\left(\frac{2 + 3\sqrt{3}}{2}\right)\cos{x}\)

\(\displaystyle \frac{\sin{x}}{\cos{x}} = \frac{-\frac{2 + 3\sqrt{3}}{2}}{\phantom{-}\frac{3 - 2\sqrt{3}}{2}}\)

\(\displaystyle \tan{x} = -\frac{2 + 3\sqrt{3}}{3 - 2\sqrt{3}}\)

\(\displaystyle \tan{x} = -\frac{(2 + 3\sqrt{3})(3 + 2\sqrt{3})}{(3 - 2\sqrt{3})(3 + 2\sqrt{3})}\)

\(\displaystyle \tan{x} = -\frac{6 + 4\sqrt{3} + 9\sqrt{3} + 18}{9 - 12}\)

\(\displaystyle \tan{x} = \frac{-(24 + 13\sqrt{3})}{-3}\)

\(\displaystyle \tan{x} = \frac{24 + 13\sqrt{3}}{3}\).
 
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Prove It

MHF Helper
Aug 2008
12,897
5,001
You can use identity \(\displaystyle \displaystyle \cos(\theta + \pi) = -\cos(\theta)\), to start out.

Edit: Well I suppose it should be \(\displaystyle \displaystyle \cos(\theta - \pi) = -\cos(\theta)\), same idea.
Yes, we coud let \(\displaystyle \theta = x + \frac{\pi}{3}\), but without the angle sum and difference identities we still can't reduce this to \(\displaystyle \tan{x}\). So refer to my post above :D
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
If 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?

I started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?

Thanks in advance.
Since Prove It posted a full solution, here's another way.

\(\displaystyle 3\sin\left(x + \dfrac{\pi}{3}\right) = 2\cos\left(x - \dfrac{2\pi}{3}\right) = -2 \cos\left(x + \dfrac{\pi}{3}\right)\)

\(\displaystyle \tan\left(x+\dfrac{\pi}{3} \right) = \dfrac{-2}{3}\)

Use

\(\displaystyle \displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)

if you know it, otherwise use sin and cos angle sum like Prove It.

Here

\(\displaystyle \dfrac{-2}{3} = \dfrac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x} \)

\(\displaystyle -2(1-\sqrt{3}\tan x) = 3(\tan x+\sqrt{3})\)

\(\displaystyle -2+2\sqrt{3}\tan x = 3\tan x+3\sqrt{3}\)

\(\displaystyle (2\sqrt{3}-3)\tan x = 3\sqrt{3}+2\)

\(\displaystyle \tan x = \dfrac{2+3\sqrt{3}}{-3+2\sqrt{3}}\)

\(\displaystyle \tan x = \dfrac{(2+3\sqrt{3})(-3-2\sqrt{3})}{(-3+2\sqrt{3})(-3-2\sqrt{3})}\)

\(\displaystyle \tan x = \dfrac{24+13\sqrt{3}}{3}\)

Yes, we coud let \(\displaystyle \theta = x + \frac{\pi}{3}\), but without the angle sum and difference identities we still can't reduce this to \(\displaystyle \tan{x}\). So refer to my post above
I was busy typing and didn't see your recent post until just now. I think your way is a bit easier than mine, anyway.