How do you solve sqrt2 sin2theta+1=0?

1 Answer
Jun 24, 2018

theta=(5pi)/8,(7pi)/8,(13pi)/8,(15pi)/8 if the domain is 0 <= theta <= 2pi

Explanation:

sqrt2 sin2theta+1=0

sqrt2 sin2theta=-1

sin2theta=-1/sqrt2

Remember that sin2theta is negative in the 3rd and 4th quadrants

2theta= pi+pi/4, 2pi-pi/4, 3pi+pi/4, 4pi-pi/4

2theta=(5pi)/4,(7pi)/4,(13pi)/4,(15pi)/4

theta=(5pi)/8,(7pi)/8,(13pi)/8,(15pi)/8

if the domain is 0 <= theta <= 2pi