You are Here: Home >< Maths

# Maths C3 Trigonometry Help

Announcements Posted on
Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. I need help in solving a maths question:

Solve this equation for x in the interval 0 < x < 2π giving your answers in terms of π.
3 sec² x = 4 tan² x.

π/3, 2π/3, 4π/3, 5π/3.

Here is my attempt (in radiance of course):

3/cos²x = 4 tan²x
3 = 4 tan²x cos²x
3/4 = tan²x cos²x
3/4= sin²x/cos²x . cos²x
3/4 =sin²x
Root 3/2 = sin x
x = sin-¹(root 3/2)
x= 1/3π
Using the CAST method, π-1/3π = 2/3 π.
Therefore, x = 1/3π, 2/3π.

Here is my second attempt, this time taking into account the equation: sec² x = 1 + tan ²x
3(1 + tan²x) = 4 tan² x
3 + 3 tan² x = 4 tan² x
4 tan² x - 3 tan² x - 3=0
tan² x - 3 =0
tan² x = 3
tan x = root 3
x = tan-¹ (root 3)
x = 1/3π.
Using the CAST method, 1/3π + π = 4/3 π
Therefore, x = 1/3π, 4/3π

If someone could identify where I went wrong it would really be appreciated.
2. (Original post by Nathan R)
I need help in solving a maths question:

Solve this equation for x in the interval 0 < x < 2π giving your answers in terms of π.
3 sec² x = 4 tan² x.

π/3, 2π/3, 4π/3, 5π/3.

Here is my attempt (in radiance of course):

3/cos²x = 4 tan²x
3 = 4 tan²x cos²x
3/4 = tan²x cos²x
3/4= sin²x/cos²x . cos²x
3/4 =sin²x
Root 3/2 = sin x
x = sin-¹(root 3/2)
x= 1/3π
Using the CAST method, π-1/3π = 2/3 π.
Therefore, x = 1/3π, 2/3π.

Here is my second attempt, this time taking into account the equation: sec² x = 1 + tan ²x
3(1 + tan²x) = 4 tan² x
3 + 3 tan² x = 4 tan² x
4 tan² x - 3 tan² x - 3=0
tan² x - 3 =0
tan² x = 3
tan x = root 3
x = tan-¹ (root 3)
x = 1/3π.
Using the CAST method, 1/3π + π = 4/3 π
Therefore, x = 1/3π, 4/3π

If someone could identify where I went wrong it would really be appreciated.
Both methods are correct, however you are missing solutions as a direct result of your square rooting. For example in the second one;

while you are taking into account the positive quantity, the negative quantity would get your the rest of the solutions. Choose one of your methods, go along with it, and consider both signs; this would get you all the solutions you need in the domain
3. (Original post by RDKGames)
Both methods are correct, however you are missing solutions as a direct result of your square rooting. For example in the second one;

while you are taking into account the positive quantity, the negative quantity would get your the rest of the solutions. Choose one of your methods, go along with it, and consider both signs; this would get you all the solutions you need in the domain
Thanks, it seems so obvious now. Your help is really appreciated.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: August 8, 2016
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:
Today on TSR

### University open days

Is it worth going? Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams