Question

cos2theta -3cos theta +2 upon sin2theta =1

Answer 1

$<marquee>☺☺Hi there☺☺</marquee>$

solve: cos2theta - 3 cos theta + 2 / sin2theta = 1

We have
cos2θ−3cosθ+2sin2θ=1
⇒cos2θ−3cosθ+2=sin2θ
⇒cos2θ−3cosθ+2=1−cos2θ
[As sin2θ=1−cos2θ]
⇒2cos2θ−3cosθ+1=0
⇒2cos2θ−2cosθ−cosθ+1=0
⇒2cosθ(cosθ−1)−1(cosθ−1)=0
⇒(2cosθ−1)(cosθ−1)=0
⇒2cosθ−1=0   or   cosθ−1=0
⇒cosθ=12   or   cosθ=1
⇒θ = 60°  or  θ = 0°(rejected)
So, θ = 60°

Thanks ♥♥

Answer 2

cos2θ−3cosθ+2sin2θ=1

cos2θ−3cosθ+2sin2θ=1⇒cos2θ−3cosθ+2=sin2θ

cos2θ−3cosθ+2sin2θ=1⇒cos2θ−3cosθ+2=sin2θ⇒cos2θ−3cosθ+2=1−cos2θ

cos2θ−3cosθ+2sin2θ=1⇒cos2θ−3cosθ+2=sin2θ⇒cos2θ−3cosθ+2=1−cos2θ[As sin2θ=1−cos2θ]

Hope its help uh :D