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Unsure about how to approach this trigonometric question

Hi All,

Another question I'm trying to complete, but don't really know how to approach:

Solve the following for -pi <x pi:

(1+2cosx)(3tan^2 x - 1) = 0
I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

Could someone please assist?
Original post by Ples#!LOL
Hi All,

Another question I'm trying to complete, but don't really know how to approach:

Solve the following for -pi <x pi:

(1+2cosx)(3tan^2 x - 1) = 0
I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

Could someone please assist?


You don't need any trig identities for this one. You have two bracketed terms multiplied together to give zero, therefore ...
Reply 2
Original post by Ples#!LOL
Hi All,

Another question I'm trying to complete, but don't really know how to approach:

Solve the following for -pi <x pi:

(1+2cosx)(3tan^2 x - 1) = 0
I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

Could someone please assist?


Since the equation is already factorised and equates to zero, you don't need to do any kind of transformation. How would you find x from (x-a)(x-b)=0? You can use the same idea here as a first step.

Although on the topic of identities for tan, there are some useful ones which you can get by starting from the classic guy:
cos2x+sin2x=1\cos^2 x + \sin^2 x =1

Divide this through by cos2xcos^2 x and you get:
1+tan2x=sec2x1 + \tan^2 x = \sec^2 x

Again though, just to reiterate, you don't need the identities here.
Reply 3
Original post by Gregorius
You don't need any trig identities for this one. You have two bracketed terms multiplied together to give zero, therefore ...


Was a bit more busy at work than expected so couldn't get back to solving the problem until now :P

Would it be cos^-1(- 1/2) = 2/3 pi, so the solutions are 2/3 pi, - 2/3pi
tan^-1(1/3) = 0.322 radians, pi - 0.322 radians.

?
Original post by Ples#!LOL
Was a bit more busy at work than expected so couldn't get back to solving the problem until now :P

Would it be cos^-1(- 1/2) = 2/3 pi, so the solutions are 2/3 pi, - 2/3pi
tan^-1(1/3) = 0.322 radians, pi - 0.322 radians.

?
To find solutions relating to the 2nd bracket, you should be solving for tan2x=13\tan^2 x = \frac{1}{3} (you appear to be solving for tanx=13\tan x = \frac{1}{3})
Reply 5
Original post by DFranklin
To find solutions relating to the 2nd bracket, you should be solving for tan2x=13\tan^2 x = \frac{1}{3} (you appear to be solving for tanx=13\tan x = \frac{1}{3})


I see! So it would be tan^2 x = 1/3,
then take x = tan^-1(√1/3)
x = 1/6pi

Then, using the Quadrant rule, 1/6 pi and 5/6 pi are solutions.
Original post by Ples#!LOL
I see! So it would be tan^2 x = 1/3,
then take x = tan^-1(√1/3)
x = 1/6pi

Then, using the Quadrant rule, 1/6 pi and 5/6 pi are solutions.


Half right, but for the wrong reasons.

Firstly:

tan5π613\tan \frac{5\pi}{6}\not= \frac{1}{\sqrt{3}}

Check which quadrants have tan positive, and what your domain is.

Secondly:

As an example.

If x2=5x^2 = 5

then x=±5 x = \pm \sqrt{5}
(edited 6 years ago)
Reply 7
Original post by ghostwalker
Half right, but for the wrong reasons.

Firstly:

tan5π613\tan \frac{5\pi}{6}\not= \frac{1}{3}

Check which quadrants have tan positive, and what your domain is.

Secondly:

As an example.

If x2=5x^2 = 5

then x=±5 x = \pm \sqrt{5}


Epic fail: I had looked at a different question previously, and thought the range was 0 <= x <= pi :P

The other solution is - 1/6pi.
Original post by Ples#!LOL
Epic fail: I had looked at a different question previously, and thought the range was 0 <= x <= pi :P

The other solution is - 1/6pi.


Getting there. -pi/6 is a solution.

5pi/6 is a solution, but not because tan x =1/root(3), rather -1/root(3)

There is one other solution you've not mentioned yet. Check the graph of tan over your domain.
Reply 9
Original post by ghostwalker
Getting there. -pi/6 is a solution.

5pi/6 is a solution, but not because tan x =1/root(3), rather -1/root(3)

There is one other solution you've not mentioned yet. Check the graph of tan over your domain.


Is -5pi/6 the final solution?
Reply 10
Original post by Ples#!LOL
Is -5pi/6 the final solution?


There is more than one solution
Reply 11
Original post by Desmos
There is more than one solution


Yes, I am aware...

That was meant as in the final solution that I have not mentioned so far :P
Original post by Ples#!LOL
Is -5pi/6 the final solution?


Yep - that's the missing one.
Reply 13
Original post by Ples#!LOL
Yes, I am aware...

That was meant as in the final solution that I have not mentioned so far :P


Oh yeah, it is.
Reply 14
Original post by ghostwalker
Yep - that's the missing one.


Original post by Desmos
Oh yeah, it is.


Thanks guys.

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