[2bon2bt=t?]

Sorry if this is a stupid question, but I really don't know how to do the Trig equations when they have more than one x, for example, how would you do this?

5sin(2x)=2cos(2x) ?


Thanks x

[A128]

(Original post by 2bon2bt=t?)
Sorry if this is a stupid question, but I really don't know how to do the Trig equations when they have more than one x, for example, how would you do this?

5sin(2x)=2cos(2x) ?


Thanks x

cant remember exactly ..look on examsolutions.co.uk under trigonometry (i think c3)

[2bon2bt=t?]

(Original post by A128)
cant remember exactly ..look on examsolutions.co.uk under trigonometry (i think c3)

Thanks, I'll have a look. Surely it can't be that hard though, this is a question from an additional maths paper and I'm still in year 11, so C3? But thanks for your help

[A128]

(Original post by 2bon2bt=t?)
Thanks, I'll have a look. Surely it can't be that hard though, this is a question from an additional maths paper and I'm still in year 11, so C3? But thanks for your help

Ooh this is GCSE, right sorry im talking about a level c3 (core maths 3) its one of the unit in A2 maths Lol...it should be there on the website somewhere, i think it has something to do with sin / cos = tan or you can try sin 2x = 2sin x but im probably wrong if its year 11 ...your welcome

[MattLBates]

sin2x / cos2x = tan 2x

See if that helps

[raheem94]

(Original post by 2bon2bt=t?)
Sorry if this is a stupid question, but I really don't know how to do the Trig equations when they have more than one x, for example, how would you do this?

5sin(2x)=2cos(2x) ?


Thanks x

Divide both sides by

Remember,

[2bon2bt=t?]

(Original post by MattLBates)
sin2x / cos2x = tan 2x

See if that helps

ohhh, thanks, I know sinx/cosx=tanx but I didn't realise how to do it with 2x.
So I've got
5sin2x=2cos2x
5tan2x=2
tan2x=2/5
inverse tan= 21.8
Now what do I do? Halve it?

[Intriguing Alias]

(Original post by 2bon2bt=t?)
ohhh, thanks, I know sinx/cosx=tanx but I didn't realise how to do it with 2x.
So I've got
5sin2x=2cos2x
5tan2x=2
tan2x=2/5
inverse tan= 21.8
Now what do I do? Halve it?

well you have 2x = 21.8 as your principal value.

What range of x are you asked to use?

[raheem94]

(Original post by 2bon2bt=t?)
ohhh, thanks, I know sinx/cosx=tanx but I didn't realise how to do it with 2x.
So I've got
5sin2x=2cos2x
5tan2x=2
tan2x=2/5
inverse tan= 21.8
Now what do I do? Halve it?

What's the given range?
Find the other solutions of , half them after finding all the solutions of

[2bon2bt=t?]

(Original post by hassi94)
well you have 2x = 21.8 as your principal value.

What range of x are you asked to use?

0<x<360

[raheem94]

(Original post by 2bon2bt=t?)
0<x<360

So the range for is

Find the other solutions of

[Intriguing Alias]

(Original post by 2bon2bt=t?)
0<x<360

If 0<x<360 then 0<2x<720

Do you know how to find the other values of 2x?

[2bon2bt=t?]

(Original post by raheem94)
So the range for is

Find the other solutions of

I got 10.9, 190.9, 370.9, and 550.9,
But the mark scheme says 2x=21.8, 201.8
so x=10.9, 190.9, 100.9, 280.9,

Thank you for your help

[raheem94]

(Original post by 2bon2bt=t?)
I got 10.9, 190.9, 370.9, and 550.9,
But the mark scheme says 2x=21.8, 201.8
so x=10.9, 190.9, 100.9, 280.9,

Thank you for your help

You have probably made some mistake while adding subtracting the angles.

[2bon2bt=t?]

(Original post by raheem94)


You have probably made some mistake while adding subtracting the angles.

right, thanks

[Intriguing Alias]

(Original post by 2bon2bt=t?)
I got 10.9, 190.9, 370.9, and 550.9,
But the mark scheme says 2x=21.8, 201.8
so x=10.9, 190.9, 100.9, 280.9,

Thank you for your help

Don't half your value of 2x yet. Find all the other values of 2x within your range and THEN half them.

[2bon2bt=t?]

(Original post by hassi94)
Don't half your value of 2x yet. Find all the other values of 2x within your range and THEN half them.

Ahhh finally got it. Thank you