Help on General Solution trig Watch

mathsRus
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Given that 4 cos x + 3 sin x = R cos(x −a) find the value of R and the value of a, where
R > 0 and 0 < a < π/2
.
(a) Hence find all values of x between 0 and 2π satisfying
i. 4 cos x + 3 sin x = 2,
ii. 4 cos 2x + 3 sin 2x = 5 cos x.
(b) Find the greatest and least values of the expression
_________1________
4 cos x + 3 sin x + 6
and give the corresponding values of x between 0 and 2π

I am stuck on part (ii) and b.

for (a)(ii) I got this far: 5cos(x - 0.6435) = 5cosx. I don't know what to do with 5cosx because it is a hence question. Do i solve the LHS by finding x?
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Notnek
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(Original post by mathsRus)
Given that 4 cos x + 3 sin x = R cos(x −a) find the value of R and the value of a, where
R > 0 and 0 < a < π/2
.
(a) Hence find all values of x between 0 and 2π satisfying
i. 4 cos x + 3 sin x = 2,
ii. 4 cos 2x + 3 sin 2x = 5 cos x.
(b) Find the greatest and least values of the expression
_________1________
4 cos x + 3 sin x + 6
and give the corresponding values of x between 0 and 2π

I am stuck on part (ii) and b.

for (a)(ii) I got this far: 5cos(x - 0.6435) = 5cosx. I don't know what to do with 5cosx because it is a hence question. Do i solve the LHS by finding x?
\displaystyle \frac{1}{4\cos x + 3\sin x +6} = \frac{1}{R\cos(x-a) + 6}

To make the fraction as small as possible, we need R\cos(x-a) to be as big as possible. What's the greatest value that \cos(x-a) can be?

Have you got an answer for aii and is it correct? You made a mistake in your working but it could have been a typo.

Edit: Do you still need help with aii?
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mathsRus
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yes please need help with aii as I thought it was just cos(x - 0.6435)
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davros
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(Original post by mathsRus)
Given that 4 cos x + 3 sin x = R cos(x −a) find the value of R and the value of a, where
R > 0 and 0 < a < π/2
.
(a) Hence find all values of x between 0 and 2π satisfying
i. 4 cos x + 3 sin x = 2,
ii. 4 cos 2x + 3 sin 2x = 5 cos x.
(b) Find the greatest and least values of the expression
_________1________
4 cos x + 3 sin x + 6
and give the corresponding values of x between 0 and 2π

I am stuck on part (ii) and b.

for (a)(ii) I got this far: 5cos(x - 0.6435) = 5cosx. I don't know what to do with 5cosx because it is a hence question. Do i solve the LHS by finding x?
Assuming your values for R and a are correct, you shouldn't have an 'x' inside the bracket on the LHS - note that the original equation in (ii) is in terms of cos2x and sin2x i.e. you've replaced x with 2x in the very first part. So your "converted" form should look like Rcos(2x - a).

Because there is a 5 on both sides you can divide by it, so now you're solving an equation like cos A = cos B and you should have a method (or methods) of finding the general solution of this type of equation.
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mathsRus
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(Original post by davros)
Assuming your values for R and a are correct, you shouldn't have an 'x' inside the bracket on the LHS - note that the original equation in (ii) is in terms of cos2x and sin2x i.e. you've replaced x with 2x in the very first part. So your "converted" form should look like Rcos(2x - a).

Because there is a 5 on both sides you can divide by it, so now you're solving an equation like cos A = cos B and you should have a method (or methods) of finding the general solution of this type of equation.
I actually don't know what to use and I did simply all the way to cos(2x-0.6435) = cosx but is there like an identity?

If I expand it as follows: cos2xcos0.6435 + sin2xsin0.6435 = cosx, it just makes it more difficult
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davros
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(Original post by mathsRus)
I actually don't know what to use and I did simply all the way to cos(2x-0.6435) = cosx but is there like an identity?

If I expand it as follows: cos2xcos0.6435 + sin2xsin0.6435 = cosx, it just makes it more difficult
Forget about expanding it!!

Isn't there a section in your book that says "this is the general solution to cos A = cos B"?

Just apply this method with A = 2x-0.6435 and B = x, then rearrange to solve for x.
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mathsRus
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(Original post by davros)
Forget about expanding it!!

Isn't there a section in your book that says "this is the general solution to cos A = cos B"?

Just apply this method with A = 2x-0.6435 and B = x, then rearrange to solve for x.
thing is we don't have a textbook and weren't taught anything like this in our lectures.
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Notnek
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(Original post by mathsRus)
thing is we don't have a textbook and weren't taught anything like this in our lectures.
If cos(A) is equal to cos(B), how could the angles A and B be related? Think about the CAST diagram or the cos graph.
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mathsRus
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(Original post by notnek)
If cos(A) is equal to cos(B), how could the angles A and B be related? Think about the CAST diagram or the cos graph.
Nope Still don't get it. Is cosA = cosB = cos(A-B)?
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davros
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(Original post by mathsRus)
Nope Still don't get it. Is cosA = cosB = cos(A-B)?
No that isn't true in general.

What level are you working at - are you studying for A levels or doing something different?
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