Question

2sin theta+ cos theta=2, find sin theta

Answer 1

Given $2 \sin \theta +\cos \theta =2$.

Rearranging the above equation and solving as a quadratic in $\sin \theta$,

$\cos \theta =2-2 \sin \theta \\ \cos^2 \theta =(2-2 \sin \theta )^2\\ 1-\sin^2 \theta=4(1-\sin \theta )^2\\ 1-\sin^2 \theta=4-8\sin \theta+4\sin^2 \theta\\ 5\sin^2 \theta-8\sin \theta+3=0 \sin \theta =\frac{8 \pm \sqrt{64-60}}{10} \\ \sin \theta =\frac{8 \pm 2}{10}$

$\sin \theta =1 \; or \; \sin \theta =0.6$