3arccosx +2arcsinx =3π/2⇔3(arccosx +arcsinx) - arcsinx = 3π/2⇔ 3π/2 - arcsinx = 3π/2
⇒x =0, следовательно (x+3)³ =(0+3)³ =27.
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4arcsinx +arccosx =π ⇔3arcsinx+ (arcsinx +arccosx) =π ⇔3arcsinx +π/2 =π⇔
arcsinx =π/6 ⇒ x =1/2 ,следовательно 3x² =3*(1/2)² =3/4.
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cos(arctq√3 +arccos√3/2) = cos(π/3 +π/6) = cosπ/2 =0.
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cos(arcsin40/41 -arcsin4/5) =cos(arcsin40/41)*cos(arcsin4/5) +sin(arcsin40/41)*sin(arcsin4/5) =cos(arccos9/41)*cos(arccos3/5) +40/41*4/5 =
9/41*3/5) +40/41*4/5 =(27+160)/5*41 = 187/205.
|a| ≤1 ,|b| ≤1 .
max(arccosa -4arccosb) = π -4(-π/2) = π +2π =3π.