Need help with this trigonometric question involving r cos(2x + a). Confused

Solve the equation 2cosx + 5sinx= 4 by expressing in the form r cos(2x+a)
* (x is actually theta and a is actually alpha)

I don't know how to start this question. Would definitely be grateful if someone could help me Thanks

Do you mean rcos(2x + a) or rcos(x + a) ?

rcos(2x+a), if it was the latter I would probably have done it with no problem. So yeah

Sorry, it's got to be a misprint unless I'm missing the blindingly obvious (always possible as my sleep has been disrupted by the storms for 2 nights running!!)

Hmmm. Because originally, the question asked us to solve it by expressing it in the form of r cos(x+a). I did that, then the next question was the one I posted. Hahha, if it's a misprint then I'm burning the book... Thank you though

Any chance you can scan / photograph the page containing the two successive questions and upload?

No worries if you can't!

Here you go
It's question number 5.
And thank you so very much

Hey, do you have the answers to it?

Well, this is kinda funny... There is no answer
I'm pretty confused hhahaha. I assumed its the same answer from question number 4 which are 26'10', 110'14'. Sorry if this doesn't make sense.
But anyway, did you solve it?

Well theta doesn't equal x. Theta's represents any angle.
That means you can solve it as usual letting 2x=theta and find values within double the range they give you: Rcos(2x+a)=Rcos(theta+a)
Then divide by 2 to find x. If that makes sense.

Here you go
It's question number 5.
And thank you so very much

I'm flummoxed I'm afraid - I can't see any way in which using 2x + a makes the equations easier, or even solvable!

The later question where it asks you to use cos(2x+a) makes more sense because you've got a cos^2 x term and a sinxcosx term so you can convert these to cos2x and sin2x and then rearrange.

I can only think question 5 is some sort of aberration on the part of the author unless your teacher has any ideas (or have you broken up now?).

How do you find r and a for the question you know how to do?

A cosx - B sinx = r cos(x + a)

Well, if we want to derive the formula, or show every single step, then you would start by expanding the cosine.

... = r cosx cosa - r sinx sina

Matching the coefficients of like terms gives A = r cosa and B = r sina

Then, by drawing the the right triangle with angle a and hypotenuse r, we find that a = arctan(B/A) and r = sqrt(A^2 + B^2).

Analogously, we probably want to solve this by expanding cos(2x + a). Then, judging by the instructions from problem 1...

Tell me if everything works out.

How do you find r and a for the question you know how to do?

A cosx - B sinx = r cos(x + a)

Well, if we want to derive the formula, or show every single step, then you would start by expanding the cosine.

... = r cosx cosa - r sinx sina

Matching the coefficients of like terms gives A = r cosa and B = r sina

Then, by drawing the the right triangle with angle a and hypotenuse r, we find that a = arctan(B/A) and r = sqrt(A^2 + B^2).

Analogously, we probably want to solve this by expanding cos(2x + a). Then, judging by the instructions from problem 1...

Tell me if everything works out.

I think we can see that - but what are you going to equate the cos2x to or the sin2x? It doesn't seem obvious where the question is leading unless we're all missing something!