Rešite 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1?

# 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1 #

# => 1 / (tan2x-tanx) -1 / (1 / (tan2x) -1 / tanx) = 1 #

# => 1 / (tan2x-tanx) + 1 / (1 / (tanx) -1 / (tan2x)) = 1 #

# => 1 / (tan2x-tanx) + (tanxtan2x) / (tan2x-tanx) = 1 #

# => (1 + tanxtan2x) / (tan2x-tanx) = 1 #

# => 1 / tan (2x-x) = 1 #

# => tan (x) = 1 = tan (pi / 4) #

# => x = npi + pi / 4 #

Odgovor:

# x = npi + pi / 4 #

Pojasnilo:

# tan2x-tanx = (sin2x) / (cos2x) -sinx / cosx = (sin2xcosx-cos2xsinx) / (cos2xcosx) #

= #sin (2x-x) / (cos2xcosx) = sinx / (cos2xcosx) #

in # cot2x-cotx = (cos2x) / (sin2x) -cosx / sinx = (sinxcos2x-cosxsin2x) / (sin2xsinx) #

= #sin (x-2x) / (sin2xsinx) = - sinx / (sin2xsinx) #

Zato # 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1 # lahko zapišemo kot

# (cos2xcosx) / sinx + (sin2xsinx) / sinx = 1 #

ali # (cos2xcosx + sin2xsinx) / sinx = 1 #

ali #cos (2x-x) / sinx = 1 #

ali # cosx / sinx = 1 # t.j. # cotx = 1 = postelja (pi / 4) #

Zato # x = npi + pi / 4 #

Kakšna je enačba črte normalne do f (x) = sec4x-cot2x pri x = pi / 3?

"Normal" => y = - (3x) / (8-24sqrt3) + (152sqrt3-120 + 3pi) / (24-72sqrt2) => y ~~ 0.089x-1.52 Normal je pravokotna črta na tangens. f (x) = sec (4x) -cot (2x) f '(x) = 4sek (4x) tan (3x) + 2csc ^ 2 (2x) f' (pi / 3) = 4sek ((4pi) / 3 ) tan ((4pi) / 3) + 2csc ^ 2 ((2pi) / 3) = (8-24sqrt3) / 3 Za normalno, m = -1 / (f '(pi / 3)) = - 3 / ( 8-24sqrt3) f (pi / 3) = sec ((4pi) / 3) -cot ((2pi) / 3) = (sqrt3-6) / 3 (sqrt3-6) / 3 = -3 / (8- 24sqrt3) (pi / 3) + cc = (sqrt3-6) / 3 + pi / (8-24sqrt3) = (152sqrt3-120 + 3pi) / (24-72sqrt2) "Normalno": y = - (3x) / (8-24sqrt3) + (152sqrt3-120 + 3pi)

Poenostavite popolnoma: 1 / cot2x - 1 / cos2x?

Rarr1 / (cot2x) -1 / (cos2x) = (sinx-cosx) / (sinx + cosx) rarr1 / (cot2x) -1 / cos2x = (sin2x) / (cos2x) -1 / (cos2x) = - (1 -2sinx * cosx) / (cos2x) = - (cos ^ 2x-2cosx * sinx + sin ^ 2x) / (cos2x) = - (cosx-sinx) ^ 2 / ((cosx + sinx) (cosx-sinx) = (sinx-cosx) / (sinx + cosx)

Roberto si deli baseball kartice enakopravno med seboj, svojim bratom in njegovimi petimi prijatelji. Roberto je ostalo 6 kart. Koliko kart je Roberto dal? Vnesite in rešite enačbo za delitev, da rešite problem. kart.

X / 7 = 6 Tako je Roberto začel s 42 kartami in dal 36. x je skupno število kart. Roberto je te karte razdelil na sedem načinov, pri čemer se je končal s šestimi kartami. 6xx7 = 42 To je skupno število kartic. Ker je obdržal 6, je dal stran 36.