Question

Question #4242c

Answer 1

Substitute `cos(2x) = 2cos^2(x)-1`
Use the quadratic formula on the resulting equation.
Use the inverse cosine function on the two cases.
Add the periodic nature of the function.

Explanation:

Given: `cos(2x)+sqrt(2)cos(x)=1`

Substitute `cos(2x) = 2cos^2(x)-1`

`2cos^2(x)-1 + sqrt2cos(x)-1 =0`

`cos^2(x) + sqrt2/2cos(x)-1 =0`

Use the quadratic formula:

`cos(x)= (-sqrt2/2+-sqrt((sqrt2/2)^2-4(1)(-1)))/(2(1))`

`cos(x)= -sqrt2/4+-3sqrt2/4`

`cos(x)= -sqrt2` and `cos(x)= sqrt2/2`

We must discard the first root, because it is outside of the range of the cosine function `-1<=cos(x)<=1`

Use the inverse cosine on the second root:

`x= cos^-1(sqrt2/2)`

We know that there are two values within the range `0<=x<2pi`

`x = pi/4` and `x = (7pi)/4`

Add the fact that these roots will repeat at integer multiples of `2pi`:

`x = pi/4+2npi` and `x = (7pi)/4+ 2npi` `n in ZZ`

These roots can be observed on the graph of the original function:

graph{cos(2x)+sqrt(2)cos(x)-1 [-10, 10, -5, 5]}