help needed for trig equations (1 Viewer)

[lisztphisto]

could sum1 please answer these questions for me, thanks in advance

Solve for x in 0<x<360

1. (cos x) ^2 = sinxcosx

2. root 3 (cosec 1/2x)^2 + cot (x/2) = root 3

3. root 3 (cosecA)^2 = 4cotA

4. (sinA)^2 - 2sinAcosA - 3(cosA)^2 = 0

THANKS A LOT

 

[pLuvia]

1. Move sinxcosx to the left hand side and factorise then solve for x
2. Change everything in terms of sinx and cosx then solve from there
3. Same as 2
4. Think about factorisation, treat sinA=x and cosA=y, then factorise and solve

 

[lisztphisto]

1. Move sinxcosx to the left hand side and factorise then solve for x
2. Change everything in terms of sinx and cosx then solve from there
3. Same as 2
4. Think about factorise, treat sinA=x and cosA=y, then factorise and solve

ok i shall have a try
ty!

 

[ianc]

yeah for the first one:

cos²x = sinx.cosx


it is tempting to just cancel a cosx from both sides, leaving you with
cosx=sinx

please DON'T do this!!!!!!!!!! Iit is a very common mistake made in trig equations......use Pluvia's method to solve it correctly.

if you want to see for yourself, first do it Pluvia's way, then try it the way I pointed out; and compare the number of solutions you get.


I hope you can see the difference between the 2 methods and why simply cancelling stuff out in trig equations will cost you lots of marks........

 

[blakwidow]

Thank you for the advice. That is very common mistake in tests

Thats a good question

-blakwidow

 

[Forbidden.]

... Move sinxcosx to the left hand side and factorise then solve for x ...

If not cos x = sin x

Then ...

cos² x = sin x . cos x
cos² x - sin x . cos x = 0

cos x (sin x - 1) = 0

cos x = 1 - sin x

(They're not even square trigonometric functions - Neither are they Pythagorean identities)

Unless 90⁰ = 1

and then x = 90⁰

...crazy ...

 

[blakwidow]

If not cos x = sin x

Then ...

cos² x = sin x . cos x
cos² x - sin x . cos x = 0

cos x (sin x - 1) = 0

cos x = 1 - sin x

(They're not even square trigonometric functions - Neither are they Pythagorean identities)

Unless 90⁰ = 1

and then x = 90⁰

...crazy ...

and after this use the trig graphs to figure out the solution IN THE GIVEN DOMAIN

 

[idling fire]

Hmm... I can only do them using the t-method. I fail. >.<

 

[pLuvia]

and after this use the trig graphs to figure out the solution IN THE GIVEN DOMAIN

The domain is the full 360 degrees, unless it's out of the domain then no need for graphs actually you wouldn't need it other wise just convert the answer to suit the domain

Hmm... I can only do them using the t-method. I fail. >.<

You better learn how to do without the t-method because you hardly ever use that method in HSC unless specified to, I remember I only used it once or twice for an exercise during my preliminary/HSC never again did I have to use it

 

[idling fire]

You better learn how to do without the t-method because you hardly ever use that method in HSC unless specified to, I remember I only used it once or twice for an exercise during my preliminary/HSC never again did I have to use it

Then I'm screwed (other than Q1).

 

[Forbidden.]

and after this use the trig graphs to figure out the solution IN THE GIVEN DOMAIN

If cos x = 1 - sin x

Then x = 90⁰, 270⁰ for 0⁰ < x < 360⁰

(ASTC thingy where cosine ratio is positive in 1st & 4th quadrants only)

Damn I never new trigonometry's dark side in 2U Maths ...