# Help on General Solution trigWatch

#1
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?
0
4 years ago
#2
(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?

To make the fraction as small as possible, we need to be as big as possible. What's the greatest value that 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?
0
#3
yes please need help with aii as I thought it was just cos(x - 0.6435)
0
4 years ago
#4
(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.
0
#5
(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
0
4 years ago
#6
(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

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.
0
#7
(Original post by davros)

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.
0
4 years ago
#8
(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.
0
#9
(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)?
0
4 years ago
#10
(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?
0
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