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# C2 Trigonometry watch

1. (sinx -1) (5cosx + 3) = 0

Solve the equation for x, in the interval 0 < x < 360

What I have done:

(sinx -1) (5cosx + 3) = 0

√cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

((1 - cosx) -1) (5cosx + 3) = 0
-cosx(5cosx + 3) = 0
-5cos^2x - 3cosx = 0
cosx(-5cosx - 3) = 0
cos x = 0 or cos x = -3/5

x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

Thank you very much!
2. (Original post by zebra015)
(sinx -1) (5cosx + 3) = 0

Solve the equation for x, in the interval 0 < x < 360

What I have done:

(sinx -1) (5cosx + 3) = 0

√cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

((1 - cosx) -1) (5cosx + 3) = 0
-cosx(5cosx + 3) = 0
-5cos^2x - 3cosx = 0
cosx(-5cosx - 3) = 0
cos x = 0 or cos x = -3/5

x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

Thank you very much!

You're making it waaaay more difficult for yourself.

(sinx -1) ---> sinx = 1, hence x = 90.
(5cosx + 3) ---> cosx = -3/5, hence x = 126.9 and (360 - 126.9 =) 233.1.

That's it. No need to substitute any identities.
3. Without going further, I can tell you that you can't square root the identity the way you did. The square root sign doesn't separate out, so instead it would look like this:

Furthermore, your equation is already factorised so you can form two equations,

4. (Original post by zebra015)
(sinx -1) (5cosx + 3) = 0

Solve the equation for x, in the interval 0 < x < 360

What I have done:

(sinx -1) (5cosx + 3) = 0

√cos^2 x + √sin^2 x = √1 - Is this step correct? I just square rooted everything in the identity - I think it's correct, but just want to make sure.

((1 - cosx) -1) (5cosx + 3) = 0
-cosx(5cosx + 3) = 0
-5cos^2x - 3cosx = 0
cosx(-5cosx - 3) = 0
cos x = 0 or cos x = -3/5

x = 90, x = 126.9 (the first values are obtained by finding the inverse cos of 0 and -3/5), x = 233.1, x = 270 (the second set of values are obtained by finding 360-x, since cos x = cos (360-x))

I'm not sure where I may have gone wrong. I did get the first 3 values of x right: x = 90, x = 126.9, x = 233.1. However, x = 270 is incorrect.

I have graphed y = (sinx -1) (5cosx + 3), and there are 3 solutions at x = 90, x = 126.9, x = 233.1.

I also graphed y = cosx(-5cosx - 3), and there are 4 solutions at x = 90, x = 126.9, x = 233.1, x = 270.

It's just that last x = 270 value that's annoying me - I can't seem to work out what's gone wrong and where it came from.

Thank you very much!
Firstly. Think of this as factorising a equation. You will get solutions by equating the factors to 0.

So to find the solutions:

Equate them to 0...
Sin x = 1 and cos x = -3/5
Solve using that as simple as that !

Also, your square root is wrong. When you square root an equation you square root both sides. You've square rooted each term individually which is wrong. √(Cos^2 x + sin^2 x) is not the same as
√cos^2 x + √sin^2 x....

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5. (Original post by zebra015)
...
Others have said it but I want to jump on this train. So, why not just say or ??
6. √( 9 + 16 ) should = 5

if you try

√9 + √16 you get 7....
7. (Original post by RDKGames)
Others have said it but I want to jump on this train. So, why not just say or ??

FML i didnt even realise it was factorised This would've made it a lot easier ..... + rep to all

thank you all for ur help!!
8. (Original post by the bear)
√( 9 + 16 ) should = 5

if you try

√9 + √16 you get 7....
Always the best . Got to rep you xD

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