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פּוֹפּוּלָרִי טריגונומטריה >

(sin^3(x))/(2+2(sin(x))^2)=1

פתרון

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

פתרון

x \in R אין פתרון ל

צעדי פתרון

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

בעזרת שיטת ההצבה

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

sin (x) = u :

נניח ש

\frac{u ^{3}}{2 + 2 u ^{2}} = 1

\frac{u ^{3}}{2 + 2 u ^{2}} = 1 : u \approx 2.35930 \dots

\frac{u ^{3}}{2 + 2 u ^{2}} = 1

2 + 2 u^{2}

הכפל את שני האגפים ב

\frac{u ^{3}}{2 + 2 u ^{2}} = 1

2 + 2 u^{2}

הכפל את שני האגפים ב

\frac{u ^{3}}{2 + 2 u ^{2}} (2 + 2 u^{2}) = 1 \cdot (2 + 2 u^{2})

פשט

\frac{u ^{3}}{2 + 2 u ^{2}} (2 + 2 u^{2}) = 1 \cdot (2 + 2 u^{2})

\frac{u ^{3}}{2 + 2 u ^{2}} (2 + 2 u^{2})

פשט את:

u^{3}

\frac{u ^{3}}{2 + 2 u ^{2}} (2 + 2 u^{2})

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

:הכפל שברים

= \frac{u ^{3} ( 2 + 2 u ^{2} )}{2 + 2 u ^{2}}

2 + 2 u^{2} :

בטל את הגורמים המשותפים

= u^{3}

1 \cdot (2 + 2 u^{2})

פשט את:

2 + 2 u^{2}

1 \cdot (2 + 2 u^{2})

1 \cdot (2 + 2 u^{2}) = (2 + 2 u^{2}) :

הכפל

= (2 + 2 u^{2})

(a) = a

:הסר סוגריים

= 2 + 2 u^{2}

u^{3} = 2 + 2 u^{2}

u^{3} = 2 + 2 u^{2}

u^{3} = 2 + 2 u^{2}

u^{3} = 2 + 2 u^{2}

פתור את:

u \approx 2.35930 \dots

u^{3} = 2 + 2 u^{2}

לצד שמאל

2 u^{2}

העבר

u^{3} = 2 + 2 u^{2}

משני האגפים

2 u^{2}

החסר

u^{3} - 2 u^{2} = 2 + 2 u^{2} - 2 u^{2}

פשט

u^{3} - 2 u^{2} = 2

u^{3} - 2 u^{2} = 2

לצד שמאל

2

העבר

u^{3} - 2 u^{2} = 2

משני האגפים

2

החסר

u^{3} - 2 u^{2} - 2 = 2 - 2

פשט

u^{3} - 2 u^{2} - 2 = 0

u^{3} - 2 u^{2} - 2 = 0

בשיטת ניטון-רפסון

u^{3} - 2 u^{2} - 2 = 0

מצא פתרון אחד ל:

u \approx 2.35930 \dots

u^{3} - 2 u^{2} - 2 = 0

הגדרת קירוב ניוטון-רפזון

f (u) = u^{3} - 2 u^{2} - 2

f^{^{'}} (u)

מצא את:

3 u^{2} - 4 u

\frac{d}{d u} (u^{3} - 2 u^{2} - 2)

(f \pm g)^{'} = f^{'} \pm g^{'}

:השתמש בחוק החיבור

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (2)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:השתמש בחוק החזקה

= 3 u^{3 - 1}

פשט

= 3 u^{2}

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

(a \cdot f)^{'} = a \cdot f^{'}

:הוצא את הקבוע

= 2 \frac{d}{d u} (u^{2})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:השתמש בחוק החזקה

= 2 \cdot 2 u^{2 - 1}

פשט

= 4 u

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

\frac{d}{d x} (a) = 0

:נגזרת של קבוע

= 0

= 3 u^{2} - 4 u - 0

פשט

= 3 u^{2} - 4 u

u_{0} = - 1

החלף

Δ u_{n + 1} < 0.000001

עד ש

u_{n + 1}

חשב

u_{1} = - 0.28571 \dots : Δ u_{1} = 0.71428 \dots

f (u_{0}) = (- 1)^{3} - 2 (- 1)^{2} - 2 = - 5

f^{^{'}} (u_{0}) = 3 (- 1)^{2} - 4 (- 1) = 7

u_{1} = - 0.28571 \dots

Δ u_{1} = ∣ - 0.28571 \dots - (- 1) ∣ = 0.71428 \dots

Δ u_{1} = 0.71428 \dots

u_{2} = 1.28991 \dots : Δ u_{2} = 1.57563 \dots

f (u_{1}) = (- 0.28571 \dots)^{3} - 2 (- 0.28571 \dots)^{2} - 2 = - 2.18658 \dots

f^{^{'}} (u_{1}) = 3 (- 0.28571 \dots)^{2} - 4 (- 0.28571 \dots) = 1.38775 \dots

u_{2} = 1.28991 \dots

Δ u_{2} = ∣ 1.28991 \dots - (- 0.28571 \dots) ∣ = 1.57563 \dots

Δ u_{2} = 1.57563 \dots

u_{3} = - 17.64595 \dots : Δ u_{3} = 18.93587 \dots

f (u_{2}) = 1.28991 \dots^{3} - 2 \cdot 1.28991 \dots^{2} - 2 = - 3.18149 \dots

f^{^{'}} (u_{2}) = 3 \cdot 1.28991 \dots^{2} - 4 \cdot 1.28991 \dots = - 0.16801 \dots

u_{3} = - 17.64595 \dots

Δ u_{3} = ∣ - 17.64595 \dots - 1.28991 \dots ∣ = 18.93587 \dots

Δ u_{3} = 18.93587 \dots

u_{4} = - 11.55537 \dots : Δ u_{4} = 6.09058 \dots

f (u_{3}) = (- 17.64595 \dots)^{3} - 2 (- 17.64595 \dots)^{2} - 2 = - 6119.35487 \dots

f^{^{'}} (u_{3}) = 3 (- 17.64595 \dots)^{2} - 4 (- 17.64595 \dots) = 1004.72330 \dots

u_{4} = - 11.55537 \dots

Δ u_{4} = ∣ - 11.55537 \dots - (- 17.64595 \dots) ∣ = 6.09058 \dots

Δ u_{4} = 6.09058 \dots

u_{5} = - 7.49987 \dots : Δ u_{5} = 4.05549 \dots

f (u_{4}) = (- 11.55537 \dots)^{3} - 2 (- 11.55537 \dots)^{2} - 2 = - 1812.00239 \dots

f^{^{'}} (u_{4}) = 3 (- 11.55537 \dots)^{2} - 4 (- 11.55537 \dots) = 446.80124 \dots

u_{5} = - 7.49987 \dots

Δ u_{5} = ∣ - 7.49987 \dots - (- 11.55537 \dots) ∣ = 4.05549 \dots

Δ u_{5} = 4.05549 \dots

u_{6} = - 4.80117 \dots : Δ u_{6} = 2.69869 \dots

f (u_{5}) = (- 7.49987 \dots)^{3} - 2 (- 7.49987 \dots)^{2} - 2 = - 536.34930 \dots

f^{^{'}} (u_{5}) = 3 (- 7.49987 \dots)^{2} - 4 (- 7.49987 \dots) = 198.74366 \dots

u_{6} = - 4.80117 \dots

Δ u_{6} = ∣ - 4.80117 \dots - (- 7.49987 \dots) ∣ = 2.69869 \dots

Δ u_{6} = 2.69869 \dots

u_{7} = - 3.00422 \dots : Δ u_{7} = 1.79694 \dots

f (u_{6}) = (- 4.80117 \dots)^{3} - 2 (- 4.80117 \dots)^{2} - 2 = - 158.77552 \dots

f^{^{'}} (u_{6}) = 3 (- 4.80117 \dots)^{2} - 4 (- 4.80117 \dots) = 88.35844 \dots

u_{7} = - 3.00422 \dots

Δ u_{7} = ∣ - 3.00422 \dots - (- 4.80117 \dots) ∣ = 1.79694 \dots

Δ u_{7} = 1.79694 \dots

u_{8} = - 1.79774 \dots : Δ u_{8} = 1.20648 \dots

f (u_{7}) = (- 3.00422 \dots)^{3} - 2 (- 3.00422 \dots)^{2} - 2 = - 47.16492 \dots

f^{^{'}} (u_{7}) = 3 (- 3.00422 \dots)^{2} - 4 (- 3.00422 \dots) = 39.09297 \dots

u_{8} = - 1.79774 \dots

Δ u_{8} = ∣ - 1.79774 \dots - (- 3.00422 \dots) ∣ = 1.20648 \dots

Δ u_{8} = 1.20648 \dots

u_{9} = - 0.95246 \dots : Δ u_{9} = 0.84527 \dots

f (u_{8}) = (- 1.79774 \dots)^{3} - 2 (- 1.79774 \dots)^{2} - 2 = - 14.27385 \dots

f^{^{'}} (u_{8}) = 3 (- 1.79774 \dots)^{2} - 4 (- 1.79774 \dots) = 16.88661 \dots

u_{9} = - 0.95246 \dots

Δ u_{9} = ∣ - 0.95246 \dots - (- 1.79774 \dots) ∣ = 0.84527 \dots

Δ u_{9} = 0.84527 \dots

u_{10} = - 0.23616 \dots : Δ u_{10} = 0.71629 \dots

f (u_{9}) = (- 0.95246 \dots)^{3} - 2 (- 0.95246 \dots)^{2} - 2 = - 4.67845 \dots

f^{^{'}} (u_{9}) = 3 (- 0.95246 \dots)^{2} - 4 (- 0.95246 \dots) = 6.53144 \dots

u_{10} = - 0.23616 \dots

Δ u_{10} = ∣ - 0.23616 \dots - (- 0.95246 \dots) ∣ = 0.71629 \dots

Δ u_{10} = 0.71629 \dots

u_{11} = 1.67454 \dots : Δ u_{11} = 1.91071 \dots

f (u_{10}) = (- 0.23616 \dots)^{3} - 2 (- 0.23616 \dots)^{2} - 2 = - 2.12472 \dots

f^{^{'}} (u_{10}) = 3 (- 0.23616 \dots)^{2} - 4 (- 0.23616 \dots) = 1.11200 \dots

u_{11} = 1.67454 \dots

Δ u_{11} = ∣ 1.67454 \dots - (- 0.23616 \dots) ∣ = 1.91071 \dots

Δ u_{11} = 1.91071 \dots

u_{12} = 3.37374 \dots : Δ u_{12} = 1.69920 \dots

f (u_{11}) = 1.67454 \dots^{3} - 2 \cdot 1.67454 \dots^{2} - 2 = - 2.91261 \dots

f^{^{'}} (u_{11}) = 3 \cdot 1.67454 \dots^{2} - 4 \cdot 1.67454 \dots = 1.71410 \dots

u_{12} = 3.37374 \dots

Δ u_{12} = ∣ 3.37374 \dots - 1.67454 \dots ∣ = 1.69920 \dots

Δ u_{12} = 1.69920 \dots

u_{13} = 2.71344 \dots : Δ u_{13} = 0.66030 \dots

f (u_{12}) = 3.37374 \dots^{3} - 2 \cdot 3.37374 \dots^{2} - 2 = 13.63622 \dots

f^{^{'}} (u_{12}) = 3 \cdot 3.37374 \dots^{2} - 4 \cdot 3.37374 \dots = 20.65152 \dots

u_{13} = 2.71344 \dots

Δ u_{13} = ∣ 2.71344 \dots - 3.37374 \dots ∣ = 0.66030 \dots

Δ u_{13} = 0.66030 \dots

u_{14} = 2.42389 \dots : Δ u_{14} = 0.28954 \dots

f (u_{13}) = 2.71344 \dots^{3} - 2 \cdot 2.71344 \dots^{2} - 2 = 3.25295 \dots

f^{^{'}} (u_{13}) = 3 \cdot 2.71344 \dots^{2} - 4 \cdot 2.71344 \dots = 11.23458 \dots

u_{14} = 2.42389 \dots

Δ u_{14} = ∣ 2.42389 \dots - 2.71344 \dots ∣ = 0.28954 \dots

Δ u_{14} = 0.28954 \dots

u_{15} = 2.36204 \dots : Δ u_{15} = 0.06185 \dots

f (u_{14}) = 2.42389 \dots^{3} - 2 \cdot 2.42389 \dots^{2} - 2 = 0.49051 \dots

f^{^{'}} (u_{14}) = 3 \cdot 2.42389 \dots^{2} - 4 \cdot 2.42389 \dots = 7.93025 \dots

u_{15} = 2.36204 \dots

Δ u_{15} = ∣ 2.36204 \dots - 2.42389 \dots ∣ = 0.06185 \dots

Δ u_{15} = 0.06185 \dots

u_{16} = 2.35930 \dots : Δ u_{16} = 0.00273 \dots

f (u_{15}) = 2.36204 \dots^{3} - 2 \cdot 2.36204 \dots^{2} - 2 = 0.01993 \dots

f^{^{'}} (u_{15}) = 3 \cdot 2.36204 \dots^{2} - 4 \cdot 2.36204 \dots = 7.28957 \dots

u_{16} = 2.35930 \dots

Δ u_{16} = ∣ 2.35930 \dots - 2.36204 \dots ∣ = 0.00273 \dots

Δ u_{16} = 0.00273 \dots

u_{17} = 2.35930 \dots : Δ u_{17} = 5.23398 E - 6

f (u_{16}) = 2.35930 \dots^{3} - 2 \cdot 2.35930 \dots^{2} - 2 = 0.00003 \dots

f^{^{'}} (u_{16}) = 3 \cdot 2.35930 \dots^{2} - 4 \cdot 2.35930 \dots = 7.26178 \dots

u_{17} = 2.35930 \dots

Δ u_{17} = ∣ 2.35930 \dots - 2.35930 \dots ∣ = 5.23398 E - 6

Δ u_{17} = 5.23398 E - 6

u_{18} = 2.35930 \dots : Δ u_{18} = 1.9156 E - 11

f (u_{17}) = 2.35930 \dots^{3} - 2 \cdot 2.35930 \dots^{2} - 2 = 1.39106 E - 10

f^{^{'}} (u_{17}) = 3 \cdot 2.35930 \dots^{2} - 4 \cdot 2.35930 \dots = 7.26173 \dots

u_{18} = 2.35930 \dots

Δ u_{18} = ∣ 2.35930 \dots - 2.35930 \dots ∣ = 1.9156 E - 11

Δ u_{18} = 1.9156 E - 11

u \approx 2.35930 \dots

הפעל חילוק ארוך:

\frac{u ^{3} - 2 u ^{2} - 2}{u - 2.35930 \dots} = u^{2} + 0.35930 \dots u + 0.84770 \dots

u^{2} + 0.35930 \dots u + 0.84770 \dots \approx 0

בשיטת ניטון-רפסון

u^{2} + 0.35930 \dots u + 0.84770 \dots = 0

מצא פתרון אחד ל:

u \in R

אין פתרון ל

u^{2} + 0.35930 \dots u + 0.84770 \dots = 0

הגדרת קירוב ניוטון-רפזון

f (u) = u^{2} + 0.35930 \dots u + 0.84770 \dots

f^{^{'}} (u)

מצא את:

2 u + 0.35930 \dots

\frac{d}{d u} (u^{2} + 0.35930 \dots u + 0.84770 \dots)

(f \pm g)^{'} = f^{'} \pm g^{'}

:השתמש בחוק החיבור

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (0.35930 \dots u) + \frac{d}{d u} (0.84770 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

:השתמש בחוק החזקה

= 2 u^{2 - 1}

פשט

= 2 u

\frac{d}{d u} (0.35930 \dots u) = 0.35930 \dots

\frac{d}{d u} (0.35930 \dots u)

(a \cdot f)^{'} = a \cdot f^{'}

:הוצא את הקבוע

= 0.35930 \dots \frac{d u}{d u}

\frac{d u}{d u} = 1

:השתמש בנגזרת הבסיסית

= 0.35930 \dots \cdot 1

פשט

= 0.35930 \dots

\frac{d}{d u} (0.84770 \dots) = 0

\frac{d}{d u} (0.84770 \dots)

\frac{d}{d x} (a) = 0

:נגזרת של קבוע

= 0

= 2 u + 0.35930 \dots + 0

פשט

= 2 u + 0.35930 \dots

u_{0} = - 2

החלף

Δ u_{n + 1} < 0.000001

עד ש

u_{n + 1}

חשב

u_{1} = - 0.86584 \dots : Δ u_{1} = 1.13415 \dots

f (u_{0}) = (- 2)^{2} + 0.35930 \dots (- 2) + 0.84770 \dots = 4.12909 \dots

f^{^{'}} (u_{0}) = 2 (- 2) + 0.35930 \dots = - 3.64069 \dots

u_{1} = - 0.86584 \dots

Δ u_{1} = ∣ - 0.86584 \dots - (- 2) ∣ = 1.13415 \dots

Δ u_{1} = 1.13415 \dots

u_{2} = 0.07141 \dots : Δ u_{2} = 0.93726 \dots

f (u_{1}) = (- 0.86584 \dots)^{2} + 0.35930 \dots (- 0.86584 \dots) + 0.84770 \dots = 1.28629 \dots

f^{^{'}} (u_{1}) = 2 (- 0.86584 \dots) + 0.35930 \dots = - 1.37239 \dots

u_{2} = 0.07141 \dots

Δ u_{2} = ∣ 0.07141 \dots - (- 0.86584 \dots) ∣ = 0.93726 \dots

Δ u_{2} = 0.93726 \dots

u_{3} = - 1.67803 \dots : Δ u_{3} = 1.74945 \dots

f (u_{2}) = 0.07141 \dots^{2} + 0.35930 \dots \cdot 0.07141 \dots + 0.84770 \dots = 0.87846 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.07141 \dots + 0.35930 \dots = 0.50213 \dots

u_{3} = - 1.67803 \dots

Δ u_{3} = ∣ - 1.67803 \dots - 0.07141 \dots ∣ = 1.74945 \dots

Δ u_{3} = 1.74945 \dots

u_{4} = - 0.65673 \dots : Δ u_{4} = 1.02129 \dots

f (u_{3}) = (- 1.67803 \dots)^{2} + 0.35930 \dots (- 1.67803 \dots) + 0.84770 \dots = 3.06057 \dots

f^{^{'}} (u_{3}) = 2 (- 1.67803 \dots) + 0.35930 \dots = - 2.99676 \dots

u_{4} = - 0.65673 \dots

Δ u_{4} = ∣ - 0.65673 \dots - (- 1.67803 \dots) ∣ = 1.02129 \dots

Δ u_{4} = 1.02129 \dots

u_{5} = 0.43640 \dots : Δ u_{5} = 1.09314 \dots

f (u_{4}) = (- 0.65673 \dots)^{2} + 0.35930 \dots (- 0.65673 \dots) + 0.84770 \dots = 1.04304 \dots

f^{^{'}} (u_{4}) = 2 (- 0.65673 \dots) + 0.35930 \dots = - 0.95417 \dots

u_{5} = 0.43640 \dots

Δ u_{5} = ∣ 0.43640 \dots - (- 0.65673 \dots) ∣ = 1.09314 \dots

Δ u_{5} = 1.09314 \dots

u_{6} = - 0.53344 \dots : Δ u_{6} = 0.96984 \dots

f (u_{5}) = 0.43640 \dots^{2} + 0.35930 \dots \cdot 0.43640 \dots + 0.84770 \dots = 1.19495 \dots

f^{^{'}} (u_{5}) = 2 \cdot 0.43640 \dots + 0.35930 \dots = 1.23210 \dots

u_{6} = - 0.53344 \dots

Δ u_{6} = ∣ - 0.53344 \dots - 0.43640 \dots ∣ = 0.96984 \dots

Δ u_{6} = 0.96984 \dots

u_{7} = 0.79587 \dots : Δ u_{7} = 1.32931 \dots

f (u_{6}) = (- 0.53344 \dots)^{2} + 0.35930 \dots (- 0.53344 \dots) + 0.84770 \dots = 0.94060 \dots

f^{^{'}} (u_{6}) = 2 (- 0.53344 \dots) + 0.35930 \dots = - 0.70758 \dots

u_{7} = 0.79587 \dots

Δ u_{7} = ∣ 0.79587 \dots - (- 0.53344 \dots) ∣ = 1.32931 \dots

Δ u_{7} = 1.32931 \dots

u_{8} = - 0.10983 \dots : Δ u_{8} = 0.90570 \dots

f (u_{7}) = 0.79587 \dots^{2} + 0.35930 \dots \cdot 0.79587 \dots + 0.84770 \dots = 1.76707 \dots

f^{^{'}} (u_{7}) = 2 \cdot 0.79587 \dots + 0.35930 \dots = 1.95104 \dots

u_{8} = - 0.10983 \dots

Δ u_{8} = ∣ - 0.10983 \dots - 0.79587 \dots ∣ = 0.90570 \dots

Δ u_{8} = 0.90570 \dots

u_{9} = - 5.98468 \dots : Δ u_{9} = 5.87485 \dots

f (u_{8}) = (- 0.10983 \dots)^{2} + 0.35930 \dots (- 0.10983 \dots) + 0.84770 \dots = 0.82030 \dots

f^{^{'}} (u_{8}) = 2 (- 0.10983 \dots) + 0.35930 \dots = 0.13963 \dots

u_{9} = - 5.98468 \dots

Δ u_{9} = ∣ - 5.98468 \dots - (- 0.10983 \dots) ∣ = 5.87485 \dots

Δ u_{9} = 5.87485 \dots

לא יכול למצוא פתרון

הפתרון למשוואה הוא

u \approx 2.35930 \dots

u \approx 2.35930 \dots

u = sin (x)

החלף בחזרה

sin (x) \approx 2.35930 \dots

sin (x) \approx 2.35930 \dots

sin (x) = 2.35930 \dots :

אין פתרון

sin (x) = 2.35930 \dots

- 1 \leq sin (x) \leq 1

אין פתרון

אחד את הפתרונות

x \in R אין פתרון ל

גרף

דוגמאות פופולריות

solvefor x,cot^2(x)= 1/3

so l v e f or x, cot^{2} (x) = \frac{1}{3}

2+cos^2(x)=-5sin(x)

2 + cos^{2} (x) = - 5 sin (x)

tan^5(x)tan^2(x)=1

tan^{5} (x) tan^{2} (x) = 1

(sin^2(a)+1)/((1+tan^2(a)))=1

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

tan(a)= 5/9 ,b=6

tan (a) = \frac{5}{9}, b = 6