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

1 Answer

x=pi/4 andx=(7pi)/4x=π4andx=7π4

Explanation:

tan (x/2) = (2 - sqrt2)/(2sin x)tan(x2)=222sinx
(sin (x/2))/(cos (x/2)) = (2 - sqrt2)/(4sin (x/2).cos (x/2))sin(x2)cos(x2)=224sin(x2).cos(x2)
Simplify by cos (x/2)cos(x2) and cross-multiply -->
4sin^2 (x/2) = 2 - sqrt24sin2(x2)=22
sin^2 (x/2) = (2 - sqrt2)/4sin2(x2)=224
sin (x/2) = +- (sqrt(2 - sqrt2))/2 = +- 0.38sin(x2)=±222=±0.38
Calculator and unit circle give 4 solutions-->
a. sin (x/2) = (sqrt(2 - sqrt2)/2) = 0.38sin(x2)=(222)=0.38 -->
x/2 = pi/8=>x=pi/4in[0,2pi]x2=π8x=π4[0,2π], and
x = 2pi-pi/4 =>x = (7pi)/4in[0,2pi]x=2ππ4x=7π4[0,2π]
b. sin( x/2) = - 0.38sin(x2)=0.38
x = - pi/4!in[0,2pi]x=π4[0,2π] and
x = - (7pi)/4!in [0,2pi]x=7π4[0,2π]
Hence,
x=pi/4 andx=(7pi)/4x=π4andx=7π4