Solve? sin6theta+sin2theta=0

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Answer 1 · 2018-05-03T04:40:52

#theta=npi/4# or #theta=(2n+1)pi/4#, where #n# is an integer

We use the formula #sinA+sinB=2sin((A+B)/2)cos((A-B)/2)#

Therefore #sin6theta+sin2theta=0#

can be written as #2sin((6theta+2theta)/2)cos((6theta-2theta)/2)=0#

or #sin4thetacos2theta=0#

Hence either #sin4theta=0# i.e. #4theta=npi# and #theta=npi/4#

or #2theta=(2n+1)pi/2# i.e. #theta=(2n+1)pi/4#

where #n# is an integer