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# CORE 3-Trigonometry watch

Solve to 3 sig. figs.:

4tan2x+3cotxsec2x=0
for 0<theta<2π
2. (Original post by Primrose96)

Solve to 3 sig. figs.:

4tan2x+3cotxsec2x=0
for 0<theta<2π
Well, we don't post solutions for people here, I'm afraid.

Clue, use trig identities to convert it all to the same trig function. That makes it easier
3. (Original post by Primrose96)

Solve to 3 sig. figs.:

4tan2x+3cotxsec2x=0
for 0<theta<2π
You may find this post gives you an idea

http://www.thestudentroom.co.uk/show....php?t=2462805
4. I have got this far. Is this correct and how can I solve it from here?
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5. (Original post by Primrose96)
I have got this far. Is this correct and how can I solve it from here?
you are on the right lines converting to tan. But you should have started by getting rid of the zero. What you have now is less helpful.

Step one is to move the 3cot x sec2x to the RHS.

then do your converting as before

you'll get a quartic in tan which can be reduced to a quadratic in tan2
6. Think I have it figured out now. Thank you so much
7. (Original post by Primrose96)
Think I have it figured out now. Thank you so much
no probs
8. Could I please have some help on another question:

Solve in radians in the range 0<x<2π

3sin2x=4cos2x

I attempted to change cos2x to (1-2(sin^2)x) but I ended up with 8(sin^2)x + 3sin2x = 4
9. (Original post by Primrose96)
Could I please have some help on another question:

Solve in radians in the range 0<x<2π

3sin2x=4cos2x

I attempted to change cos2x to (1-2(sin^2)x) but I ended up with 8(sin^2)x + 3sin2x = 4
remember tan is just sin/cos. That simplifies life considerably
10. Just finished this question. Haha I was getting worried coz I got stuck half way through. Anyway I'll give you a hint:

Turn it all into tan. Then there will be a formula that can be factorised into double brackets
11. (Original post by kesupile)
Just finished this question. Haha I was getting worried coz I got stuck half way through. Anyway I'll give you a hint:

Turn it all into tan. Then there will be a formula that can be factorised into double brackets
No. It's simpler than that
12. (Original post by Plato's Trousers)
No. It's simpler than that
Haha dude I'm talking about the problem that OP posted

And I agree with you. The second one just requires you to remember that tan(x) is sin(x)/cos(x)
13. (Original post by kesupile)
Haha dude I'm talking about the problem that OP posted

And I agree with you. The second one just requires you to remember that tan(x) is sin(x)/cos(x)
Oh right. Sorry. Yes, the original problem is just a quadratic
14. Am I along the right lines?

3sin2x = 4cos2x

let 2x=y

3siny=4cosy

siny/cosy = 4/3

tany = 4/3

tan(2x) = 4/3
15. (Original post by Primrose96)
Am I along the right lines?

3sin2x = 4cos2x

let 2x=y

3siny=4cosy

siny/cosy = 4/3

tany = 4/3

tan(2x) = 4/3
Yes. If we want to be really pedantic, then you should let , not let , but you are entirely correct in your logic.
16. (Original post by Primrose96)
Am I along the right lines?

3sin2x = 4cos2x

let 2x=y

3siny=4cosy

siny/cosy = 4/3

tany = 4/3

tan(2x) = 4/3
spot on
17. Prove the identities:

cos2A = (1-tan2A) / (1+tan2A)

For this is it better to work with the LHS or RHS?
18. (Original post by Primrose96)
Prove the identities:

cos2A = (1-tan2A) / (1+tan2A)

For this is it better to work with the LHS or RHS?
I'd start with the RHS and write tan in terms of sin and cos
19. Solve in the range 0<x<360

tan2x - tanx = 0

So far I have done 2tanx / (1-(tan^2)x) = tanx

2tanx = tanx (1 - (tan^2)x)

2 = 1 - ((tan^2)x)

(tan^2)x = -1

Where am I going wrong as you can't have a square root of a minus number?
20. (Original post by Primrose96)
Solve in the range 0<x<360

tan2x - tanx = 0

So far I have done 2tanx / (1-(tan^2)x) = tanx

2tanx = tanx (1 - (tan^2)x)

2 = 1 - ((tan^2)x)

(tan^2)x = -1

Where am I going wrong as you can't have a square root of a minus number?
Ok, this is driving me mad.

It's clearly wrong since the answer is 0 and 180, but I'm damned if I can see your error!

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