How do you solve #sqrt2 sin2theta+1=0#?

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Answer 1 · 2018-06-24T04:33:24

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

#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#