General Solution of an Equation

[babybells96]

Find the general solution of this equation:3 tan x- (square root) 3 = 0. To start this problem, I added square root 3 from both sides and then divided body sides by 3, to get tan x = (square root) 3 over 3. I know that tangent is opposite over adjacent, thus, I know that the negative square root three must be across from theta. I do not think this problem is not a 30-60-90 or a 45-45-90 triangle, so I used the pyth theorem to solve for hypotenuse and got 3 square root 2. Now, I am not quite sure how to solve this problem. Thank you!

 

[srmichael]

Find the general solution of this equation:3 tan x- (square root) 3 = 0. To start this problem, I subtracted square root 3 from both sides and then divided body sides by 3, to get tan x = - (square root) 3 over 3. I know that tangent is opposite over adjacent, thus, I know that the negative square root three must be across from theta. I do not think this problem is not a 30-60-90 or a 45-45-90 triangle, so I used the pyth theorem to solve for hypotenuse and got 3 square root 2. Now, I am not quite sure how to solve this problem. Thank you!

Not quite.

$\displaystyle 3tanx-\sqrt{3}=0$

$\displaystyle \ Now \ ADD \sqrt{3} \ to \ both \ sides:$

$\displaystyle 3tanx=\sqrt{3}$

$\displaystyle tanx=\frac{\sqrt{3}}{3}$

$\displaystyle x=tan^{-1}(\frac{\sqrt{3}}{3})$

$\displaystyle x=\frac{\pi}{6}$

 

[babybells96]

Thank you so much for fixing my mistake and pointing me in the right direction. I did not understand that I needed to use the tangent inverse to get my answer. It makes sense now! Thank you.

 

[MarkFL]

$\displaystyle \dfrac{\pi}{6}$ is a special angle for which you should know the values of the trig. functions there.

Since you asked for a general solution and $\displaystyle \tan(x+k\pi)=\tan(x)$ where $\displaystyle k\in\mathbb{Z}$, then your general solution is:

$\displaystyle x=\dfrac{\pi}{6}+k\pi=\dfrac{\pi}{6}(6k+1)$