Solve the equation (√3)sin x = cos x, in the interval 0 < x < 360
I literally have no idea how to go about doing this. This is what I've done so far:
(√3)sin x = cos x
3sin^2 x  cos^2 x = 0
3sin^2 x  (1  sin^2 x) = 0
4sin^2 x = 1
sin x = 1/2
x = 30, 150
Not sure whether I'm on the right lines or not. The correct answers are 30 & 210.
Thanks for any help.

zebra015
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 07022017 18:03

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 07022017 18:04
(Original post by zebra015)
Solve the equation (√3)sin x = cos x, in the interval 0 < x < 360
I literally have no idea how to go about doing this. This is what I've done so far:
(√3)sin x = cos x
3sin^2 x  cos^2 x = 0
3sin^2 x  (1  sin^2 x) = 0
4sin^2 x = 1
sin x = 1/2
x = 30, 150
Not sure whether I'm on the right lines or not. The correct answers are 30 & 210.
Thanks for any help.
Does that help? 
zebra015
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 07022017 18:14
I'm self teaching maths from the edexcel books and there don't seem to be any examples where 2 trig functions are involved. This is the first question I'm attempting of this kind.
I'm not really sure how I could use sin x / cos x =tan x in this case, since tan isn't involved?
thanks for your help btw 
TheFirstOrder
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 07022017 18:16
(Original post by zebra015)
I'm OK at solving simple equations with 1 trig function, e.g. 7sin x = 3. However when there are 2 trig functions, I just get confused.
I'm self teaching maths from the edexcel books and there don't seem to be any examples where 2 trig functions are involved. This is the first question I'm attempting of this kind.
I'm not really sure how I could use sin x / cos x =tan x in this case, since tan isn't involved?
thanks for your help btw 
Science Finger
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 07022017 18:17
Try (√3)sin x = cos x
(√3)(sin x /cosx) = 1
(√3)tah x = 1
tanx= (1/(√3)) 
zebra015
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 07022017 18:30
(Original post by Science Finger)
Try (√3)sin x = cos x
(√3)(sin x /cosx) = 1
(√3)tah x = 1
tanx= (1/(√3))
However could someone please explain why my method is wrong and this is right? I would've never guessed to do it this way ..... thank you!! 
Chittesh14
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 07022017 20:32
(Original post by zebra015)
Solve the equation (√3)sin x = cos x, in the interval 0 < x < 360
I literally have no idea how to go about doing this. This is what I've done so far:
(√3)sin x = cos x
3sin^2 x  cos^2 x = 0
3sin^2 x  (1  sin^2 x) = 0
4sin^2 x = 1
sin x = 1/2
x = 30, 150
Not sure whether I'm on the right lines or not. The correct answers are 30 & 210.
Thanks for any help.
EDIT: you can't square an equation like that. You'll get extra solutions as a result. In order to see the correct ones you should substitute them into the original equation and eliminate the ones that don't match it. In this case, 150 degrees is eliminated. Try it and you'll see. Also if you know your graphs well you'd know that cos 150 is ve whereas root3*sin 150 is +ve so u already know it's wrong.
Posted from TSR MobileLast edited by Chittesh14; 07022017 at 20:35. 
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 07022017 20:39
(Original post by zebra015)
Solve the equation (√3)sin x = cos x, in the interval 0 < x < 360
I literally have no idea how to go about doing this. This is what I've done so far:
(√3)sin x = cos x
3sin^2 x  cos^2 x = 0
3sin^2 x  (1  sin^2 x) = 0
4sin^2 x = 1
sin x = 1/2
x = 30, 150
Not sure whether I'm on the right lines or not. The correct answers are 30 & 210.
Thanks for any help.
Sin x = +/ 1/2, you got rid of the negative sin x which is wrong because that's how you'd get your 210 degrees lol. X = 30 when sin x = 1/2 if you draw quadrant diagram it is also 210 degrees or 330 degrees. You can now test in the original equation and eliminate 330 degrees. You will get 4 solutions instead of 2 due to squaring the equation. I hope you understand my point
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zebra015
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 07022017 20:48
(Original post by Chittesh14)
Your method doesn't flow. Can you explain how you got from one line to the other, because it's definitely wrong but we need to understand what you were thinking and then we can tell you why you're wrong.
EDIT: you can't square an equation like that. You'll get extra solutions as a result. In order to see the correct ones you should substitute them into the original equation and eliminate the ones that don't match it. In this case, 150 degrees is eliminated. Try it and you'll see. Also if you know your graphs well you'd know that cos 150 is ve whereas root3*sin 150 is +ve so u already know it's wrong.
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(√3)sin x = cos x (original equation)
3sin^2 x = cos^2 x (square everything)
3sin^2 x  cos^2 x = 0 (take away cos^2 x from both sides)
3sin^2 x  (1  sin^2 x) = 0 (replace cos^2 x with 1  sin^2 x, since cos^2 x + cos^2 x = 1)
3sin^2 x  1 + sin^2 x = 0 (multiply (1  sin^2 x) out by 1)
4sin^2 x = 1 (rearranging)
sin x = 1/2 (square root everything)
sin x = 1/2
x = 30, 150
the other method is a lot easier and I understand how it works now, however I would appreciate it if you could find the flaw in my method, as I certainly cannot thank you!Last edited by zebra015; 07022017 at 21:43. 
Chittesh14
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 07022017 20:52
(Original post by zebra015)
Sure, sorry if it was a bit vague.
(√3)sin x = cos x (original equation)
3sin^2 x = cos^2 x (square everything)
3sin^2 x  cos^2 x = 0 (take away cos^2 x from both sides)
3sin^2 x  (1  sin^2 x) = 0 (replace cos^2 x with 1  sin^2 x, since cos^2 x + cos^2 x = 1)
3sin^2 x  1 + sin^2 x = 0 (multiply (1  sin^2 x) out by 1)
4sin^2 x = 1 (rearranging)
sin x = 1/2 (square root everything)
sin x = 1/2
x = 30, 150
the other method is a lot easier and I understand how it works now, however I would appreciate it if you could find the flaw in my method, as I certainly cannot thank you!
You're welcome .
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zebra015
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 07022017 22:03
(Original post by Chittesh14)
No problem and sorry I edited my post once I realised what you did as you saw and if you could read that  I mentioned your flaw/error. Read both of my posts again and absorb the information in there because it is essential and I've made similar mistakes in the past and they're extremely harmful.
You're welcome .
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Definitely helped me a lot! 
zebra015
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 07022017 22:11
(Original post by Chittesh14)
No problem and sorry I edited my post once I realised what you did as you saw and if you could read that  I mentioned your flaw/error. Read both of my posts again and absorb the information in there because it is essential and I've made similar mistakes in the past and they're extremely harmful.
You're welcome .
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tan x = tan x (2 + 3 sin x)
1 = (2 + 3 sin x) (dividing whole equation through by tan x  I think this may be the source of error, but I can't see why this would be wrong)
1 = (2 + 3 sin x)
3 sin x = 1
x = 199.5, 340.5
The above two values are correct, however I seem to be missing 180 and 360  this may be something to do with me cancelling out tanx, however it felt like the logical thing to do :/ Thank you!! 
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 07022017 22:23
(Original post by zebra015)
Also, if you wouldn't mind, could you quickly help me with this last question?
tan x = tan x (2 + 3 sin x)
1 = (2 + 3 sin x) (dividing whole equation through by tan x  I think this may be the source of error, but I can't see why this would be wrong)
1 = (2 + 3 sin x)
3 sin x = 1
x = 199.5, 340.5
The above two values are correct, however I seem to be missing 180 and 360  this may be something to do with me cancelling out tanx, however it felt like the logical thing to do :/ Thank you!!
Can you post the full question ? The range must be > 0 so ... 0 < x <= 360 degrees.
In these questions what id do is expand the brackets and rearrange etc. If u can't solve then u must divide as long as it doesn't lose a solution or use other identities to try to solve. Never divide straight away esp when you see tan x is a factor because then it shows it's a solution.
Posted from TSR MobileLast edited by Chittesh14; 07022017 at 22:34. 
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 07022017 22:35
Hint: expand and rearrange. You'll see a common factor of tan x, now think how to solve .
Posted from TSR MobileLast edited by Chittesh14; 07022017 at 22:36. 
zebra015
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 07022017 23:48
(Original post by Chittesh14)
No problem, read the post above for help.
Hint: expand and rearrange. You'll see a common factor of tan x, now think how to solve .
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Thank you so much! Wish I could repay you somehow for your help!
I read that if you divide through by a variable, it has to be a nonzero variable, otherwise a solution will be lost ... otherwise, it should be OK. However from now on I will definitely be expanding and simplifying!!
thanks again 
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 07022017 23:50
(Original post by zebra015)
Thank you so much! Wish I could repay you somehow for your help!
I read that if you divide through by a variable, it has to be a nonzero variable, otherwise a solution will be lost ... otherwise, it should be OK. However from now on I will definitely be expanding and simplifying!!
thanks again 
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 07022017 23:54
(Original post by Chittesh14)
Yeah, it must be that lol I was trying to think but couldn't remember. And no problem, just keep posting questions  that's practice for me so it helps ! You have to expand and simplify all the time and use identities  that's the key to success!
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