How do I solve tan(x/2)=(2-sqrt(2))/2sinx for x, with restrictions of an interval [0,2pi] ?

The actual answer is pi/4 and 7pi/4; but other than that no information is provided.

1 Answer
maganbhai P.
Apr 3, 2018

#x=pi/4 and (7pi)/4#
Note:
#color(red)((1)sintheta=2sin(theta/2)cos(theta/2)#
#color(blue)((2)costheta=1-2sin^2(theta/2)#

Explanation:

Here,
#Tan(x/2)=(2-sqrt2)/(2sinx)#
#=>sinxtan(x/2)=(2-sqrt2)/2#
#=>color(red)(2sin(x/2)cos(x/2))(sin(x/2)/cos(x/2))=1-1/sqrt2#
#=>2sin^2(x/2)=1-1/sqrt2#
#=>2sin^2(x/2)-1=-1/sqrt2#
#=>color(blue)(1-2sin^2(x/2))=1/sqrt2#
#=>cosx=1/sqrt2 > 0 => I^(st)# quadrant #andIV^(th)#quadrant
But, we have ,
#x in[0,2pi)=>x=pi/4 and (7pi)/4#

Related questions
Impact of this question
1002 views around the world
You can reuse this answer
Creative Commons License