How do you solve trigonometric equations by factoring?

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Answer 1 · 2014-11-05T23:43:45

Example

Find #theta# in #[0,2pi)# such that #2sin^2theta+sin theta-1=0#.

Since

#2x^2+x-1=(2x-1)(x+1)#,

by replacing #x# by #sin theta#,

#2sin^2theta+sin theta-1=(2sin theta-1)(sin theta+1)=0#

#=>{(sin theta=1/2 => theta=pi/6"," {5pi}/6),(sin theta=-1 => theta={3pi}/2):}#

Hence, the solutions are #theta=pi/6, {5pi}/6, {3pi}/2#.

I hope that this was helpful.