Question

How do I find all solutions for `2cos^2x-1=0` on `[0,2pi]`?

Answer 1

`x=1/4pi, 3/4pi`

Explanation:

`2cos^2x-1 = 0`

`2cos^2x= 1`

`cos^2x=1/2`

`cosx=+-sqrt2/2`

`x=1/4pi, 3/4pi`

Answer 2

`x=1/4pi, 3/4pi,5/4pi,7/4pi`

Explanation:

`2cos^2x-1 = 0`

`2cos^2x= 1`

`cos^2x=1/2`

`cosx=+-sqrt2/2`

`x=1/4pi, 3/4pi,5/4pi,7/4pi`

Answer 3

Factor using the pattern `a^2-b^2=(a+b)(a-b)`
Solve each factor for x, when the factor equals 0.

Explanation:

Given: `2cos^2(x) – 1 = 0`

Factor using the pattern `a^2-b^2=(a+b)(a-b)`

`(sqrt2cos(x)+1)(sqrt2cos(x)-1)= 0`

Solve each factor for the condition that makes it become 0:

`cos(x)=-1/sqrt2 and cos(x) = 1/sqrt2`

Rationalize the fractions:

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

The values of x are well known for these conditions:

`x = pi/4,3pi/4,5pi/4 and 7pi/4`