How do you solve #\sin ^ { 2} \theta = 3( \cos ( - \theta ) - 1)#?

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Answer 1 · 2017-06-02T09:01:17

#theta=2npi#, where #n# is an integer.

As #sin^2theta=1-cos^2theta# and #cos(-theta)=costheta#

#sin^2theta=3(cos(-theta)-1)# can be written as

#1-cos^2theta=3(costheta-1)#

or #1-cos^2theta=3costheta-3#

or #cos^2theta+3costheta-3-1=0#

or #cos^2theta+3costheta-4=0#

or #cos^2theta-costheta+4costheta-4=0#

or #costheta(costheta-1)+4(costheta-1)=0#

or #(costheta+4)(costheta-1)=0#

As #costheta+4!=0#, we must have #costheta-1=0#

or #costheta=1#

and #theta=2npi#, where #n# is an integer.