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

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

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

and

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

So in your case

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

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

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

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

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

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

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

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

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

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

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

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

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 aboveYou 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.

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

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

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

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

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

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

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

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

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

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

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

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.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