Question

How do you solve `2-cos^2(x)=4sin^2(1/2x)` in the interval [0, 2pi]?

Answer 1

`pi/4, (3pi)/4, (5pi)/4 and (7pi)/4`

Explanation:

Replace in the equation `cos^2 x` by `( 1 - sin^2 x)` and put the equation in standard form:
`f(x) = 2 - 1 + sin^2 x - 4sin^2 (x/2) = 0`
Replace `sin^2 x` by `(4sin^2 (x/2).cos^2 (x/2))`
`f(x) = 1 + 4sin^2 (x/2).cos^2 (x/2) - 4sin^2 (x/2) = 0`
`f(x) = 1 + 4sin^2 (x/2)(cos^2 (x/2) - 1) = 0`
Since `(cos^2 (x/2) - 1) = - sin^2 (x/2)`, therefor:
`f(x) = 1 - 4sin^4 (x/2) = 0`
`4sin^4 (x/2) = 1`
`sin^4 (x/2) = 1/4` -->
`sin^2 (x/2) = 1/2` --> `sin x = +- 1/sqrt2 = +- sqrt2/2`
Trig table and unit circle -->
a. `sin x = sqrt2/2`. There are 2 solution arcs:
`x = pi/4` and `x = (3pi)/4`
b. `sin x = -sqrt2/2`. Two solution arcs:
`x = (5pi)/4` and `x = (7pi)/4`