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

1 Answer
Nghi N · maganbhai P.
Apr 3, 2018

#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#

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