# Thread: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

1. ## Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Stuck on this question, Thanks for your help.

2. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Originally Posted by UrgedTaylor
Stuck on this question, Thanks for your help.
When you have an equation equal to zero,
like this one,
the simplest strategy is to find a common term and write factors.
Then use the fact that zero multiplied by anything is zero.

3. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

You can factor out $\displaystyle \cos(x)$ to give $\displaystyle \cos(x)(\sqrt{3}\tan(x) + 1) = 0$

Hence either $\displaystyle \cos(x) = 0 \text{ or } \sqrt{3}\tan(x)+1=0$

4. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Thanks both of you, but i am still having issues. I'm doing a practice provincial exam for Pmath 12 and I haven't done this stuff in half a year ( also the unit I struggled the most with).

Here are the possible answer's:
A- (pi)/6 and 7(pi)/6
B-5(pi)/6 and 11(pi/6
C- (pi)/6 and 7(pi)/6 and (pi)/2 and 3(pi)/2
D-5(pi)/6 and 11(pi/6 and (pi)/2 and 3(pi)/2

Thanks for any further help

5. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Which of the answers look correct to you,
based on post 3 ?

6. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

I'm sure it will be either C or D, but that's about all I can tell so far because I know cos(pi)/2 and cos 3(pi)/2 =0 and I know there will be 2 more answers

7. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Originally Posted by UrgedTaylor
I'm sure it will be either C or D, but that's about all I can tell so far because I know cos(pi)/2 and cos 3(pi)/2 =0 and I know there will be 2 more answers
Yes, the other answers come from the 2nd factor.
You need to rearrange that for tanx, so that you can find the angles
associated with tanx.

8. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Figured out it is D Because √3tanx is negative 1 in quadrants 3 and 4. Thank you again

9. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

\displaystyle \begin{aligned} \sqrt{3}\cos{x}\tan{x}+\cos{x} & = \sqrt{3}\sin{x}+\cos{x} \\&= 2\bigg(\frac{\sqrt{3}}{2}\sin{x}+\frac{1}{2}\cos{x }\bigg) \\& = 2\bigg(\cos{\frac{\pi}{6}}~\sin{x}+\sin{\frac{\pi} {6}}\cos{x}\bigg) \\& = 2\sin\left(x+\frac{\pi}{6}\right).\end{aligned}

10. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Well, it's actually negative in quadrants 2 and 4,
because tanx gives the slope of a line through the origin.

11. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Do you know which values of $\displaystyle \displaystyle x$ make $\displaystyle \displaystyle \tan{x} = \frac{1}{\sqrt{3}}$?

12. ## Re: Solve √3 cosxtanx+ cosx=0 , where 0≤x<2(Pi)

Originally Posted by Prove It
Do you know which values of $\displaystyle \displaystyle x$ make $\displaystyle \displaystyle \tan{x} = \frac{1}{\sqrt{3}}$?
$\displaystyle \pi/6 \bmod \pi$ ?