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# Higher Maths watch

1. Could anyone help me with this question?

Solve algebraically the equation 5sin2x = 10cosx for 0 < or equal to x < 360
2. (Original post by Sthat)
Could anyone help me with this question?

Solve algebraically the equation 5sin2x = 10cosx for 0 < or equal to x < 360
You need to use:

3. (Original post by Sthat)
Could anyone help me with this question?

Solve algebraically the equation 5sin2x = 10cosx for 0 < or equal to x < 360
As soon as you see , you should think about using an identity.

Have you tried this? Please post any working that you've done.
4. (Original post by notnek)
As soon as you see , you should think about using an identity.

Have you tried this? Please post any working that you've done.
I have not, I don't really know how to start answering it.
5. (Original post by Sthat)
I have not, I don't really know how to start answering it.
SherlockHolmes gave you the identity that you should be using. Have you seen this identity before?

So start by replacing with .
6. (Original post by notnek)
SherlockHolmes gave you the identity that you should be using. Have you seen this identity before?

So start by replacing with .
Okay so after doing that it becomes 10sinxcosx = 10cos x
I then rearranged it to 10sinxcosx - 10cosx = 0 and then cosx(10sinx-10) = 0

Is that right?
7. (Original post by Sthat)
Okay so after doing that it becomes 10sinxcosx = 10cos x
I then rearranged it to 10sinxcosx - 10cosx = 0 and then cosx(10sinx-10) = 0

Is that right?
Yes that's correct. Can you carry on from there?
8. (Original post by notnek)
Yes that's correct. Can you carry on from there?
There's more? haha woops I thought that was the end.
9. This is A Level maths, right?
10. (Original post by Carthaginian)
This is A Level maths, right?
Nope, the Scottish Equivalent.

The proper definition would be SQA Higher CFE Maths.
11. (Original post by MathsAndCoffee)
Nope, the Scottish Equivalent.

The proper definition would be SQA Higher CFE Maths.
Ah, thought this was GCSE level and got scared asf.
12. (Original post by Sthat)
There's more? haha woops I thought that was the end.
You need to solve it to find values of x that satisfy the equation.

Now you've factorised the next step is to solve it like you would a quadratic with two brackets:

( ... )( ... ) = 0
13. (Original post by Carthaginian)
Ah, thought this was GCSE level and got scared asf.
Our GCSE is national 5 / intermediate 2.

Higher is A1

Correct me if I'm wrong.
14. (Original post by notnek)
You need to solve it to find values of x that satisfy the equation.

Now you've factorised the next step is to solve it like you would a quadratic with two brackets:

( ... )( ... ) = 0
Would you be able to help me with this question?

15. (Original post by Sthat)
Would you be able to help me with this question?

you need to integrate 4cos(2x) with respect to x, using limits 0 and π/4
16. (Original post by Sthat)
Would you be able to help me with this question?

Integral of 4cos2xdx = 2sin2x.
Substitute in the limits between pi/4, 0.
(2sin2(pi/4) - (2sin2(0)
(2sin(90)) - 0
sin90= 1

2x1= 2units squared!

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Updated: April 16, 2017
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