We have to prove that (pi/6)(π6) and (5pi)/65π6 are the 2 roots of the equation:
2cos^2 (pi + x) + 2sin (pi + x) = 02cos2(π+x)+2sin(π+x)=0
First, replace in the equation x by pi/6, or x = 30^@π6,orx=30∘, then, evaluate the equation.
2cos^2 (pi + pi/6) = 2cos^2 ((7pi)/6) = 2cos^2 (210)) = 1.502cos2(π+π6)=2cos2(7π6)=2cos2(210))=1.50
3sin ((7pi)/6) = 3sin (210) = 3(-0.5) = - 1.503sin(7π6)=3sin(210)=3(−0.5)=−1.50. Proved.
Next, replace x by (5pi)/6, or x = 150^@5π6,orx=150∘:
2cos^2 ((11pi)/6) = 2cos^2 (330) = 2(0.75) = 1.502cos2(11π6)=2cos2(330)=2(0.75)=1.50
3sin ((11pi)/6) = 3sin (330) = 3(-0.5) = - 1.503sin(11π6)=3sin(330)=3(−0.5)=−1.50. Proved