## adnanchowdhury Group Title Trigonometric Identities question: http://d.pr/i/FsnP I got my answer to be 3/2 but the mark scheme says -3/2. 2 years ago 2 years ago

$3\cos2 \beta+19\cos \beta+13=0$ $Use ~~~~\cos2\beta=2\cos^2\beta-1$ $3(2\cos^2\beta-1)+19\cos \beta+13=0$ $6\cos^2\beta-3+19\cos \beta+13=0$ $6\cos^2\beta+19\cos \beta+10=0$ $(2\cos \beta+5)(3\cos \beta+2)=0$ $\cos \beta=-5/2, \cos \beta =-2/3$