Question

6 cos2x + cosx-1=0

Answer 1

$6cos2x+cosx-1=0\\ 6 cdot (cos^{2}x-sin^{2}x)+cosx-1=0\\ 6cdot (cos^{2}x-1+cos^{2}x)+cosx-1=0\\ 6cdot (2cos^{2}x-1)+cosx-1=0\\ 12cos^{2}x+cosx-7=0\\ cosx=t;\\ 12t^{2}+t-7=0\\ D=1-4*(-7)*12=337\\ t=frac{-1+-sqrt{337}}{24}\\ cos x=frac{-1+-sqrt{337}}{24}\\ x=arccosfrac{-1+-sqrt{337}}{24}$