1. ## trigonometric equations

I can't seem to do (a)(ii) of this question, there are probably two ways to do it but i get stuck using my way.
Given $4cos\theta+3sin\theta=Rcos(\theta-\alpha)$ find the value of E and the value of $\alpha$ where R>0 and 0< $\alpha<\frac{\pi}{2}$
[I got R = 5 and [tex]\alpha=0.6435 rads which is right]

Hence ind all values between 0 and 2 $\pi$ satisfying
(a)(i) 4cos $\theta+3sin\theta=2$
[i also got this right its 1.8 and 5.77 rads]
(ii) 4cos $2\theta+3sin2\theta=5cos\theta$
i expanded again and got 5cos( $(2\theta-0.64350)=5cos\theta$ but i dont know where to go from here,,,ive come across a similar scenario before and cant remember what I did, If anybody could help that'd be great, thnx

2. If two angles α and β (between 0 and 2π) have the same cosine then either α = β or α + β = 2π.

3. hmm but the answer givwen is 0.21,0.64,2.31 and 4.4 rads how did u get at least two of them exactly,,,do you use sina +sinb=2sin(a+b/2)cos(a+b/2) or is that possible?'', oh i see what you mean now but that only gives me 0.6435 ans 0.231

4. You know that $5\cos(2\theta - 0.64350) = 5\cos\theta$. Cancel the 5s, then you see that you have two angles with the same cosine. there are then four possibilities:

1. $2\theta - 0.6435 = \theta$.

2. $2\theta - 0.6435 + \theta = 2\pi \approx 6.2632$.

3. $2\theta - 0.6435 + \theta = 0$.

4. $2\theta - 0.6435 + \theta = 4\pi \approx 12.5264$.

You could add any multiple of 2π to the right-hand side of 1. or 2., but the only ways to get solutions for θ in the interval from 0 to 2π are the four given above.

5. thanks!! I get the right answer now, Is that some rule to remember i.e for the limit of 2 $\pi$ you can only go to twice i.e 4pi or is there some kinda proof i could search in the internet thanks!?