Soloving sqrt(5)Cos(2x + 0.464)

1. Jun 5, 2008

thomas49th

1. The problem statement, all variables and given/known data

Solve for the interval 0 < x < pi/2

$$\sqrt{5}Cos(2x+0.464)$$

2. Relevant equations

3. The attempt at a solution

Am i right in saying as cos oscillate between -1 and 1:

The maximum is sqrt(5) and occurs when (2x-0.464) = 1

Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36

How do you solve equations like above :S

Thanks

yes my calculator is in radians.

2. Jun 5, 2008

nicksauce

FIrst of all, what is it that you are trying to solve?

3. Jun 5, 2008

thomas49th

oops i forgot.

solve for x in the range stated above

Thanks ;)

4. Jun 5, 2008

HallsofIvy

Staff Emeritus
In order to solve any equation, you first have to have an equation. What equation are you talking about. Exactly what is the question you want to answer?

Yes, the maximum of the expression you give is sqrt(5) but it does not occur when 2x+ 0.464= 1, it occurs when 2x+0.464= 0 or a multiplie of 2pi which is, I presume, where you got the values of 0.464/2 and (2pi- 0.463)/2 (it should be -0.464/2= -0.232 and (2pi 0.463)/2). But what is the question to which your book says the answer is 0.32 and 2.36?

5. Jun 5, 2008

thomas49th

oops i forgot.

solve for x in the range stated above

Thanks ;)

6. Jun 5, 2008

thomas49th

The rest of the question though is:

I) Express 2Cos2x-Sin2x in the form RCos(2x+a) giving R exactly and a to 3dp
II) Hence find the x-coordinates of the points of interestction of C1 and C2 in the interval 0<=x=<pi. Give your answer as radians and to 2dp

Note that c1 and c2 are curves with the equations y = cos2x - 2sin²x and y = sin2x respectively.

7. Jun 5, 2008

thomas49th

it' ok i got it. cos-1(1/sqrt(5))

easy peasy really :)