# SOLVEDTrigonometric equation?

#### D7236

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?

#### undefined

MHF Hall of Honor
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?

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
You should know

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

and

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

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

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

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

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

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

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

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

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

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

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

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

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

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#### Prove It

MHF Helper
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

#### undefined

MHF Hall of Honor
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?

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.