Question

How to solve for x when restricted to interval [0,2pi] tan(x/2)=(2-sqrt(2))/2sinx?

Answer 1

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

Explanation:

`tan (x/2) = (2 - sqrt2)/(2sin x)`
`(sin (x/2))/(cos (x/2)) = (2 - sqrt2)/(4sin (x/2).cos (x/2))`
Simplify by `cos (x/2)` and cross-multiply -->
`4sin^2 (x/2) = 2 - sqrt2`
`sin^2 (x/2) = (2 - sqrt2)/4`
`sin (x/2) = +- (sqrt(2 - sqrt2))/2 = +- 0.38`
Calculator and unit circle give 4 solutions-->
a. `sin (x/2) = (sqrt(2 - sqrt2)/2) = 0.38` -->
`x/2 = pi/8=>x=pi/4in[0,2pi]`, and
`x = 2pi-pi/4 =>x = (7pi)/4in[0,2pi]`
b. `sin( x/2) = - 0.38`
`x = - pi/4!in[0,2pi]` and
`x = - (7pi)/4!in [0,2pi]`
Hence,
`x=pi/4 andx=(7pi)/4`