How do you solve trigonometric equations by factoring?

1 Answer
Nov 5, 2014

Example

Find thetaθ in [0,2pi)[0,2π) such that 2sin^2theta+sin theta-1=02sin2θ+sinθ1=0.

Since

2x^2+x-1=(2x-1)(x+1)2x2+x1=(2x1)(x+1),

by replacing xx by sin thetasinθ,

2sin^2theta+sin theta-1=(2sin theta-1)(sin theta+1)=02sin2θ+sinθ1=(2sinθ1)(sinθ+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.