How do you find the general solution of sqrt2 sin2theta+1=0?

2 Answers
Jun 24, 2018

theta=-pi/8+npi or theta=-(3pi)/8+npi where n is an integer

Explanation:

sqrt2sin2theta+1=0

sqrt2sin2theta=-1

sin2theta=-1/sqrt2

Recall that sin2theta is negative in both the 3rd and 4th quadrant

2theta=-pi/4,-(3pi)/4,...

2theta=-pi/4+2npi or 2theta=-(3pi)/4+2npi where n is an integer

theta=-pi/8+npi or theta=-(3pi)/8+npi

Jun 24, 2018

The solutions are S={5/8pi+kpi, 7/8pi+kpi,}, k in ZZ

Explanation:

The equation is

sqrt2sin2theta+1=0

=>, sqrt2sin2theta=-1

=>, sin2theta=-1/sqrt2

Therefore,

{(2theta=5/4pi+2kpi),(2theta=7/4pi+2kpi),(2theta=13/4pi+2kpi),(2theta=15/4pi+2kpi):}

<=>, {(theta=5/8pi+kpi),(theta=7/8pi+kpi),(theta=13/8pi+kpi),(theta=15/8pi+kpi):}, AA k in ZZ