CORE 3-Trigonometry
Please please help with the following:
Solve to 3 sig. figs.:
4tan2x+3cotxsec2x=0
Solve to 3 sig. figs.:
4tan2x+3cotxsec2x=0
for 0<theta<2π
Scroll to see replies
Original post by Primrose96
Please please help with the following:
Solve to 3 sig. figs.:
4tan2x+3cotxsec2x=0
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
Original post by Primrose96
Please please help with the following:
Solve to 3 sig. figs.:
4tan2x+3cotxsec2x=0
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/showthread.php?t=2462805
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
(edited 10 years ago)
Think I have it figured out now. Thank you so much 
Original post by Primrose96
Think I have it figured out now. Thank you so much
no probs
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
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
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
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
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
Turn it all into tan. Then there will be a formula that can be factorised into double brackets
No. It's simpler than that
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)
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)
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
Am I along the right lines?
3sin2x = 4cos2x
let 2x=y
3siny=4cosy
siny/cosy = 4/3
tany = 4/3
tan(2x) = 4/3
3sin2x = 4cos2x
let 2x=y
3siny=4cosy
siny/cosy = 4/3
tany = 4/3
tan(2x) = 4/3
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
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.
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
3sin2x = 4cos2x
let 2x=y
3siny=4cosy
siny/cosy = 4/3
tany = 4/3
tan(2x) = 4/3
spot on
Prove the identities:
cos2A = (1-tan2A) / (1+tan2A)
For this is it better to work with the LHS or RHS?
cos2A = (1-tan2A) / (1+tan2A)
For this is it better to work with the LHS or RHS?
Original post by Primrose96
Prove the identities:
cos2A = (1-tan2A) / (1+tan2A)
For this is it better to work with the LHS or RHS?
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
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?
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?
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?
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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