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

## Homework Statement

Solve for the interval 0 < x < pi/2

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

## 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.

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

oops i forgot.

solve for x in the range stated above

Thanks ;)

HallsofIvy
Homework Helper

## Homework Statement

Solve for the interval 0 < x < pi/2

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

## 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.
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?

oops i forgot.

solve for x in the range stated above

Thanks ;)

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.

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

easy peasy really :)