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

**UrgedTaylor** · June 19th 2011, 02:19 PM

Stuck on this question, Thanks for your help.

**Archie Meade** · June 19th 2011, 02:24 PM

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.

**e^(i*pi)** · June 19th 2011, 02:24 PM

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

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

**UrgedTaylor** · June 19th 2011, 02:34 PM

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

**Archie Meade** · June 19th 2011, 02:45 PM

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

**UrgedTaylor** · June 19th 2011, 02:50 PM

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

**Archie Meade** · June 19th 2011, 02:54 PM

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.

**UrgedTaylor** · June 19th 2011, 03:35 PM

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

**TheCoffeeMachine** · June 19th 2011, 03:56 PM

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

**Archie Meade** · June 19th 2011, 04:00 PM

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

**Prove It** · June 19th 2011, 09:12 PM

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

**Moo** · June 20th 2011, 12:24 AM

Originally Posted by Prove It

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

$\pi/6 \bmod \pi$ ?

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

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

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

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

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

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