c3 trig QUESTION HELP PLZ

Pure T

Solve to one dp
5tan^2 2x - 13 sec 2x = 1

for x in interval 0-x-360

Please guys cheers :smile:

Reply 1

yusufu

tan²2x = sec²2x -1

rewrite and solve as quadratic equation in sec2x.

Reply 2

Pure T

yeh just noticed that so does it become 4 sec^2 2x - 13 sec2x =1

or 5 sec^2 2x -1 - 13sec 2x =1

Reply 3

Pure T

no soz duh lol its the 2nd one but i still cant factorise it, can u?

Reply 4

Dez

I make it out as (subbing y = sec 2x for simplicity):
5y² - 5 - 13y = 1
5y² - 13y - 6 = 0
Which factorises to:
(x - 2)(5x - 3) = 0
Fairly easy from there.

Reply 5

Pure T

yeh but where did ur random -5 come from in ur first line? am i being dumb?

Reply 6

Dez

Okay, original equation is:
5tan²2x - 13sec2x = 1
Substitute tan²2x = sec²2x - 1
5(sec²2x - 1) - 13sec2x = 1
5sec²2x - 5 - 13sec2x = 1

Does that make more sense? The 5 comes from multiplying out that bracket.

Reply 7

Pure T

ohh ffs lol cheers! where u in sussex im in hove?

Reply 8

Dez

Portslade, and I'm studying at BHASVIC, if that's your next question :P

Reply 9

Pure T

Haha lol nope! U in last year then? What u get in ur as maths? u can help me with another one though if u want to lol

ln(2x^2 +3x-9) + 2+ln(2x^2 - 7x +6) give answer in terms of e?

I think im on right tracks but got no idea tbh without the answers!

Reply 10

Dez

I'm in my last year, I got an A in AS Maths, and do you need to simplify that, or is it an equation?

Reply 11

Pure T

well just says solve in terms of e?

Reply 12

Dez

You can't solve it if it doesn't equal something. Are you sure that's all it says?

Reply 13

Pure T

oh soz mt bad thats all it says but for part a u had to simplify the first brackets divided by the second brackets and so getting (x+3)/(x-2)

therefore its ln (x+3)= 2+ln (x-2)

solve in terms of e??? thats all i got

Reply 14

Dekota

steffroud

its ln (x+3)= 2+ln (x-2)
solve in terms of e???

ln|x+3|= 2 + ln|x-2|
ln|x+3|- ln|x-2| = 2
ln|(x+3)/(x-2)|= 2 (*using the laws of logs)
(x+3)/(x-2) = e^2
(x+3) = e^2(x-2)
x+3 = xe^2 - 2e^2
xe^2 - x = 2e^2 + 3
x(e^2 - 1) = 2e^2 + 3

so; x = (2e^2 + 3)/(e^2 - 1)

Reply 15

Dez

Sinister...

ln(x+3) = 2 + ln(x-2)
ln(x+3) - ln(x-2) = 2
ln((x+3)/(x-2)) = 2 [Some law of logs]
e² = (x+3) / (x-2)
e²(x-2) = (x+3)
e²x - 2e² = x+3
e²x - x = 2e² + 3
(e²-1)x = 2e² + 3
x = (2e²+3) / (e²-1)

That's not a nice question, I might have screwed up a calculation along the line, but hopefully that's right. If you were still feeling up to it, you could plug that into your calculator and get a value, but since it asks for an answer in terms of e, that should do it.

Edit: Beaten to the mark, but at least I'm right :P

Reply 16

Pure T

cheers ! thank u! no its one of those one from solomon press, if u havent dont them they are a bit harder than the actual edexcel ones!

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