<prompt> Tafuta pembe zote theta, 0 le theta le 2 pi, yenye sifa ifuatayo: Kwa nambari zote halisi x, 0 le x le 1,x2 cos theta - x1 - x 1 - x2 sin theta 0.  <response>: Tukichukua x 0, tunapata sin theta 0. Tukichukua x 1, tunapata cos theta 0. Kwa hivyo, 0 theta fracpi 2.Kisha tunaweza kuandikaanzaalignx2 cos theta - x1 - x 1 - x2 dhambi theta x2 cos theta - 2x 1 - x sqrtcos theta sin theta 1 - x2 sin theta 2x 1 - x sqrt cos theta sin theta - x1 - x x sqrtcos theta - 1 - x sqrtsin theta2 x1 - x 2 sqrtcos theta sin theta - 1.endalignKutatua x sqrtcos theta 1 - x sqrtsin theta, tunapatax fracsqrtsin thetasqrtcos theta sqrtsin theta,ambayo iko katika muda wa 0,1 . Kwa thamani hii ya x, usemi unakuwax1 - x 2 sqrtcos theta sin theta - 1,ambayo inalazimisha 2 sqrtcos theta sin theta - 1 0, au 4 cos theta sin theta 1. Sawa, sin 2 theta frac12. Tangu 0 theta fracpi2, 0 2 theta pi, na suluhisho ni fracpi6 2 theta frac5 pi6, aufracpi12 theta frac5 pi12.Kinyume chake, ikiwa fracpi12 theta frac5 pi12, kisha cos theta 0, sin theta 0, na sin 2 theta frac12, soanzaalignx2 cos theta - x1 - x 1 - x2 dhambi theta x2 cos theta - 2x 1 - x sqrtcos theta sin theta 1 - x2 sin theta 2x 1 - x sqrt cos theta sin theta - x1 - x x sqrtcos theta - 1 - x sqrtsin theta2 x1 - x 2 sqrtcos theta sin theta - 1 0.endalignKwa hivyo, suluhu theta ni theta in left fracpi12, frac5 pi12 kulia. 