The Student Room Group

Trigonometry equations

4sin(x+30)=tan(x+30). i made tan to be sin/cos (x+30) but im not sure how to get x from here
Original post by izzy21234
4sin(x+30)=tan(x+30). i made tan to be sin/cos (x+30) but im not sure how to get x from here

Mult both sides by cos

Move everything to one side

Factorise
Reply 2
Original post by RDKGames
Mult both sides by cos

Move everything to one side

Factorise

thank you for replying but im still quite confused. i got 4sin(x+30)-(sin/cos)(x+30)=0.
in terms of factorising i got sin(x+30)(4-1/cos)=0. what do i do from here?
Original post by izzy21234
thank you for replying but im still quite confused. i got 4sin(x+30)-(sin/cos)(x+30)=0.
in terms of factorising i got sin(x+30)(4-1/cos)=0. what do i do from here?

If uv=0, we have u=0 or v=0, so...

EDIT: Just in case, I assume you didn't write the equation exactly as you've typed (when I say exactly, I do mean it), because it doesn't make sense.
(edited 9 months ago)
Reply 4
Original post by tonyiptony
If uv=0, we have u=0 or v=0, so...

EDIT: Just in case, I assume you didn't write the equation exactly as you've typed (when I say exactly, I do mean it), because it doesn't make sense.

so do i do sin(|x+30)=0 and 4-(1/cos)=0 as thats what i tried and i still cant get the answers
Original post by izzy21234
so do i do sin(|x+30)=0 and 4-(1/cos)=0 as thats what i tried and i still cant get the answers


I suggest you write/type your equations properly.
Currently, 4-(1/cos)=0 doesn't make sense.

Another suggestion, back to the very beginning, you see how everything inside the trig functions are x+30? Maybe letting y=x+30 would make your life easier...

EDIT: Also, you say you "can't get the answers". Well, what did you get? Showing your incorrect/incomplete work would be quite helpful. If typing your work seems too much work, taking a picture also works.
(edited 9 months ago)
Reply 6
IMG_2553.jpegsorry i didn’t realise i could add photos! does this make more sense as to what i’ve done?
Original post by tonyiptony
I suggest you write/type your equations properly.
Currently, 4-(1/cos)=0 doesn't make sense.

Another suggestion, back to the very beginning, you see how everything inside the trig functions are x+30? Maybe letting y=x+30 would make your life easier...

EDIT: Also, you say you "can't get the answers". Well, what did you get? Showing your incorrect/incomplete work would be quite helpful. If typing your work seems too much work, taking a picture also works.
Original post by izzy21234
IMG_2553.jpegsorry i didn’t realise i could add photos! does this make more sense as to what i’ve done?


Okay, much clearer now, thanks. You are almost there, but a few comments.

When RDK says tan(x+30) = sin/cos(x+30), they mean sin(x+30)/cos(x+30). The previous expression doesn't make sense verbatim, so that might be why you are missing the whole other branch of solutions (though the idea is correct). That's what I mean writing things properly.

Now the branch with sin(x+30) = 0. It's not the case here that if x=-30 is a solution then x=180+30 is also a solution. This deduction is made when you make the step of eliminating sin, not at the end. So the correct argument would be "there is a solution when x+30=0 implies there is also a solution when x+30=180-0".

That said, two suggestions to make your life easier:
(i) Let y=x+30 at the very beginning (like the very first line, perhaps after copying the question), then y=0 is a solution also means y=180-0=180 is also a solution; or
(ii) You can really read off the solution from this equation. Hopefully you know x+30 = 0, 180, 360, etc. So...

Also, what is the range of x the question is requiring? Be wary of that as well.
(edited 9 months ago)
Reply 8
Original post by tonyiptony
Okay, much clearer now, thanks. You are almost there, but a few comments.

When RDK says tan(x+30) = sin/cos(x+30), they mean sin(x+30)/cos(x+30). The previous expression doesn't make sense verbatim, so that might be why you are missing the whole other branch of solutions (though the idea is correct). That's what I mean writing things properly.

Now the branch with sin(x+30) = 0. It's not the case here that if x=-30 is a solution then x=180+30 is also a solution. This deduction is made when you make the step of eliminating sin, not at the end. So the correct argument would be "there is a solution when x+30=0 implies there is also a solution when x+30=180-0".

That said, two suggestions to make your life easier:
(i) Let y=x+30 at the very beginning (like the very first line, perhaps after copying the question), then y=0 is a solution also means y=180-0=180 is also a solution; or
(ii) You can really read off the solution from this equation. Hopefully you know x+30 = 0, 180, 360, etc. So...

Also, what is the range of x the question is requiring? Be wary of that as well.

thank you so much this was really useful and now i know i can add photos to make it all a lot easier!!!

Quick Reply

Latest