You are Here: Home >< Maths

# Unsure about how to approach this trigonometric question watch

1. 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.

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.

You don't need any trig identities for this one. You have two bracketed terms multiplied together to give zero, therefore ...
3. (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.

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:

Divide this through by and you get:

Again though, just to reiterate, you don't need the identities here.
4. (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

?
5. (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

?
To find solutions relating to the 2nd bracket, you should be solving for (you appear to be solving for )
6. (Original post by DFranklin)
To find solutions relating to the 2nd bracket, you should be solving for (you appear to be solving for )
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.
7. (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:

Secondly:

As an example.

If

then
8. (Original post by ghostwalker)
Half right, but for the wrong reasons.

Firstly:

Secondly:

As an example.

If

then
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.
9. (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.
10. (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?
11. (Original post by Ples#!LOL)
Is -5pi/6 the final solution?
There is more than one solution
12. (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
13. (Original post by Ples#!LOL)
Is -5pi/6 the final solution?
Yep - that's the missing one.
14. (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.
15. (Original post by ghostwalker)
Yep - that's the missing one.
(Original post by Desmos)
Oh yeah, it is.
Thanks guys.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: September 19, 2017
Today on TSR

### Results day under a month away

How are you feeling?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams