Question #bf3a5

1 Answer
Somebody N.
Feb 22, 2018

#theta=(3pi)/4+2pik#

#theta=-(3pi)/4+2pik#

Explanation:

#2cos(theta)+sqrt(2)=0#

#2cos(theta)=-sqrt(2)#

#cos(theta)=(-sqrt(2))/2#

#theta=arccos(cos(theta))=arccos(-sqrt(2)/2)=>theta=(3pi)/4+2pik#

#theta=arccos(cos(theta))=-arccos(-sqrt(2)/2)=>theta=-(3pi)/4+2pik#

For any integer #bbk#

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