bl64
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solve these equations giving values f x between 0 and 360 inclusive

a) 2sec2x - cot2x = tan2x
b) sec x + tan x = 0
c) 2tanx/2 + 3tanx = 0

for a) I changed the equation into 2/cos2x - cos2x/sin2x = sin2x/cos2x but didn't know how to make that into an easy expression

for b) I made secx 1/cosx and tanx sinx/cosx and times the whole thing by cos x to get 1 + sinx = 0 but apparently there are no solutions according to the answer, so could someone please explain why

For c) i made x/2 = y so 2tany + 3tan2y = 0 but didn;t know what to do there

can someone please help
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TeeEm
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(Original post by bl64)
solve these equations giving values f x between 0 and 360 inclusive

a) 2sec2x - cot2x = tan2x
b) sec x + tan x = 0
c) 2tanx/2 + 3tanx = 0

for a) I changed the equation into 2/cos2x - cos2x/sin2x = sin2x/cos2x but didn't know how to make that into an easy expression

for b) I made secx 1/cosx and tanx sinx/cosx and times the whole thing by cos x to get 1 + sinx = 0 but apparently there are no solutions according to the answer, so could someone please explain why

For c) i made x/2 = y so 2tany + 3tan2y = 0 but didn;t know what to do there

can someone please help
for a) after you switch into sines and cosines

multiply by cos2x
then by sin2x
and then I hope you remeber your Y12 trig identities
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weasdown
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For the first one multiply through by sinx cosx and see where it takes you.

I'm not sure what's going on with the second one.

Don't let the x/2 in the last one lead you astray. Instead of x/2 and x, replace the with x and 2x and go from there.

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bl64
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(Original post by TeeEm)
for a) after you switch into sines and cosines

multiply by cos2x
then by sin2x
and then I hope you remeber your Y12 trig identities
I am in year 12, we skipped C2 trig
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TenOfThem
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(Original post by bl64)
for b) I made secx 1/cosx and tanx sinx/cosx and times the whole thing by cos x to get 1 + sinx = 0 but apparently there are no solutions according to the answer, so could someone please explain why
You are at the point where you have Sin(x) = -1

Do you know what Cos(x) would be for that angle?

This should help you see why there are no solutions
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TeeEm
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(Original post by bl64)
solve these equations giving values f x between 0 and 360 inclusive

a) 2sec2x - cot2x = tan2x
b) sec x + tan x = 0
c) 2tanx/2 + 3tanx = 0

for a) I changed the equation into 2/cos2x - cos2x/sin2x = sin2x/cos2x but didn't know how to make that into an easy expression

for b) I made secx 1/cosx and tanx sinx/cosx and times the whole thing by cos x to get 1 + sinx = 0 but apparently there are no solutions according to the answer, so could someone please explain why

For c) i made x/2 = y so 2tany + 3tan2y = 0 but didn;t know what to do there

can someone please help
for b there are no solutions

sinx = -1

has solutions when cosx =0, so your denominator

Easy to see from graphs.
secx+ tanx = 0
secx = -tanx

both graphs share asymtotes
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Gaiaphage
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(Original post by weasdown)
Instead of x/2 and x, replace the[m] with x and 2x and go from there.
I'd advise against this, replacing x/2 with y is a good idea. If you replace it with x then you may forget to halve it, and you/an examiner may get confused with what's your new x and what's the x they're asking for

(Original post by bl64)
for b) I made secx 1/cosx and tanx sinx/cosx and times the whole thing by cos x to get 1 + sinx = 0 but apparently there are no solutions according to the answer, so could someone please explain why

can someone please help
Never divide/multiply by a variable if it then removes it from the equation - you'll lose (or in this case gain) solutions
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TeeEm
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(Original post by bl64)
I am in year 12, we skipped C2 trig
then use the identity
Cos2A + sin2A = 1
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TenOfThem
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(Original post by Gaiaphage)
I'd advise against this, replacing x/2 with y is a good idea. If you replace it with x then you may forget to halve it, and you/an examiner may get confused with what's your new x and what's the x they're asking for
Well Said
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bl64
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(Original post by TeeEm)
for a) after you switch into sines and cosines

multiply by cos2x
then by sin2x
and then I hope you remeber your Y12 trig identities
if I multiply by cos 2x and sin2x I get cos^2(2x) and sin^2(2x) what can I do with that?
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TeeEm
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(Original post by bl64)
if I multiply by cos 2x and sin2x I get cos^2(2x) and sin^2(2x) what can I do with that?
cos2A + sin2A = 1

change everything into sines

you get a quadratic in sine I think
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