<prompt> Tatuaarcsin x arcsin 2x fracpi3.  <response>: Kutoka kwa equation iliyotolewa,arcsin 2x fracpi3 - arcsin x.Kishadhambi arcsin 2x sin left fracpi3 - arcsin x kulia.Kwa hivyo, kutoka kwa fomula ya kutoa pembe,anzaalign2x sin fracpi3 cos arcsin x - cos fracpi3 sin arcsin x fracsqrt32 cdot sqrt1 - x2 - fracx2.endalignKisha 5x sqrt3 cdot sqrt1 - x2. Kuweka pande zote mbili, tunapata25x2 3 - 3x2,hivyo 28x2 3. Hii inasababisha x pm fracsqrt2114.Ikiwa x -fracsqrt2114, basi arcsin x na arcsin 2x ni hasi, kwa hivyo x -fracsqrt21 14 si suluhu.Kwa upande mwingine, 0 fracsqrt2114 frac12, so0 arcsin fracsqrt2114 fracpi6.Pia, 0 fracsqrt217 frac 1sqrt2, hivyo0 arcsin fracsqrt217 fracpi4.Kwa hiyo,0 arcsin fracsqrt2114 arcsin fracsqrt217 frac5 pi12.Pia,anzaalignsin left arcsin fracsqrt2114 arcsin fracsqrt217 kulia fracsqrt2114 cos kushoto arcsin fracsqrt217 kulia cos left arcsin fracsqrt2114 kulia cdot frac sqrt217 fracsqrt2114 cdot sqrt1 - frac2149 sqrt1 - frac21196 cdot frac sqrt217 fracsqrt32.endalignTunahitimisha hivyoarcsin fracsqrt2114 arcsin fracsqrt217 fracpi3.Kwa hivyo, suluhu pekee ni x fracsqrt2114. 