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פּוֹפּוּלָרִי >

(cos(2x))^2=cos(x)

פתרון

(cos (2 x))^{2} = cos (x)

פתרון

x = 0 + 2 πn, x = 2 π - 0 + 2 πn, x = 1.13774 \dots + 2 πn, x = 2 π - 1.13774 \dots + 2 πn

מעלות

x = 0^{\circ} + 36 0^{\circ} n, x = 36 0^{\circ} + 36 0^{\circ} n, x = 65.18792 \dots^{\circ} + 36 0^{\circ} n, x = 294.81207 \dots^{\circ} + 36 0^{\circ} n

צעדי פתרון

(cos (2 x))^{2} = cos (x)

משני האגפים

cos (x)

החסר

cos^{2} (2 x) - cos (x) = 0

Rewrite using trig identities

cos^{2} (2 x) - cos (x)

cos (2 x) = 2 cos^{2} (x) - 1

:הפעל זהות של זווית כפולה

= (2 cos^{2} (x) - 1)^{2} - cos (x)

(- 1 + 2 cos^{2} (x))^{2} - cos (x) = 0

בעזרת שיטת ההצבה

(- 1 + 2 cos^{2} (x))^{2} - cos (x) = 0

cos (x) = u :

נניח ש

(- 1 + 2 u^{2})^{2} - u = 0

(- 1 + 2 u^{2})^{2} - u = 0 : u \approx 1.00000 \dots, u \approx 0.41964 \dots

(- 1 + 2 u^{2})^{2} - u = 0

(- 1 + 2 u^{2})^{2} - u

הרחב את:

1 - 4 u^{2} + 4 u^{4} - u

(- 1 + 2 u^{2})^{2} - u

(- 1 + 2 u^{2})^{2} : 1 - 4 u^{2} + 4 u^{4}

(a + b)^{2} = a^{2} + 2 ab + b^{2}

:הפעל נוסחת הכפל המקוצר

a = - 1, b = 2 u^{2}

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

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

פשט את:

1 - 4 u^{2} + 4 u^{4}

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

(- a) = - a

:הסר סוגריים

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

(- 1)^{2} = 1

(- 1)^{2}

זוגי n

אם

, (- a)^{n} = a^{n}

:הפעל את חוק החזקות

(- 1)^{2} = 1^{2}

= 1^{2}

1^{a} = 1

הפעל את החוק

= 1

2 \cdot 1 \cdot 2 u^{2} = 4 u^{2}

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

2 \cdot 1 \cdot 2 = 4 :

הכפל את המספרים

= 4 u^{2}

(2 u^{2})^{2} = 4 u^{4}

(2 u^{2})^{2}

(a \cdot b)^{n} = a^{n} b^{n}

:הפעל את חוק החזקות

= 2^{2} (u^{2})^{2}

(u^{2})^{2} : u^{4}

(a^{b})^{c} = a^{b c}

:הפעל את חוק החזקות

= u^{2 \cdot 2}

2 \cdot 2 = 4 :

הכפל את המספרים

= u^{4}

= 2^{2} u^{4}

2^{2} = 4

= 4 u^{4}

= 1 - 4 u^{2} + 4 u^{4}

= 1 - 4 u^{2} + 4 u^{4}

= 1 - 4 u^{2} + 4 u^{4} - u

1 - 4 u^{2} + 4 u^{4} - u = 0

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

כתוב בצורה הסטנדרטית

4 u^{4} - 4 u^{2} - u + 1 = 0

בשיטת ניטון-רפסון

4 u^{4} - 4 u^{2} - u + 1 = 0

מצא פתרון אחד ל:

u \approx 1.00000 \dots

4 u^{4} - 4 u^{2} - u + 1 = 0

הגדרת קירוב ניוטון-רפזון

f (u) = 4 u^{4} - 4 u^{2} - u + 1

f^{^{'}} (u)

מצא את:

16 u^{3} - 8 u - 1

\frac{d}{d u} (4 u^{4} - 4 u^{2} - u + 1)

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

:השתמש בחוק החיבור

= \frac{d}{d u} (4 u^{4}) - \frac{d}{d u} (4 u^{2}) - \frac{d u}{d u} + \frac{d}{d u} (1)

\frac{d}{d u} (4 u^{4}) = 16 u^{3}

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

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

:הוצא את הקבוע

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

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

:השתמש בחוק החזקה

= 4 \cdot 4 u^{4 - 1}

פשט

= 16 u^{3}

\frac{d}{d u} (4 u^{2}) = 8 u

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

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

:הוצא את הקבוע

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

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

:השתמש בחוק החזקה

= 4 \cdot 2 u^{2 - 1}

פשט

= 8 u

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

\frac{d u}{d u}

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

:השתמש בנגזרת הבסיסית

= 1

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

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

:נגזרת של קבוע

= 0

= 16 u^{3} - 8 u - 1 + 0

פשט

= 16 u^{3} - 8 u - 1

u_{0} = 1

החלף

Δ u_{n + 1} < 0.000001

עד ש

u_{n + 1}

חשב

u_{1} = 1 : Δ u_{1} = 0

f (u_{0}) = 4 \cdot 1^{4} - 4 \cdot 1^{2} - 1 + 1 = 0

f^{^{'}} (u_{0}) = 16 \cdot 1^{3} - 8 \cdot 1 - 1 = 7

u_{1} = 1

Δ u_{1} = ∣ 1 - 1 ∣ = 0

Δ u_{1} = 0

u \approx 1.00000 \dots

הפעל חילוק ארוך:

\frac{4 u ^{4} - 4 u ^{2} - u + 1}{u - 1} = 4 u^{3} + 4 u^{2} - 1

4 u^{3} + 4 u^{2} - 1 \approx 0

בשיטת ניטון-רפסון

4 u^{3} + 4 u^{2} - 1 = 0

מצא פתרון אחד ל:

u \approx 0.41964 \dots

4 u^{3} + 4 u^{2} - 1 = 0

הגדרת קירוב ניוטון-רפזון

f (u) = 4 u^{3} + 4 u^{2} - 1

f^{^{'}} (u)

מצא את:

12 u^{2} + 8 u

\frac{d}{d u} (4 u^{3} + 4 u^{2} - 1)

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

:השתמש בחוק החיבור

= \frac{d}{d u} (4 u^{3}) + \frac{d}{d u} (4 u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (4 u^{3}) = 12 u^{2}

\frac{d}{d u} (4 u^{3})

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

:הוצא את הקבוע

= 4 \frac{d}{d u} (u^{3})

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

:השתמש בחוק החזקה

= 4 \cdot 3 u^{3 - 1}

פשט

= 12 u^{2}

\frac{d}{d u} (4 u^{2}) = 8 u

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

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

:הוצא את הקבוע

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

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

:השתמש בחוק החזקה

= 4 \cdot 2 u^{2 - 1}

פשט

= 8 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

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

:נגזרת של קבוע

= 0

= 12 u^{2} + 8 u - 0

פשט

= 12 u^{2} + 8 u

u_{0} = 1

החלף

Δ u_{n + 1} < 0.000001

עד ש

u_{n + 1}

חשב

u_{1} = 0.65 : Δ u_{1} = 0.35

f (u_{0}) = 4 \cdot 1^{3} + 4 \cdot 1^{2} - 1 = 7

f^{^{'}} (u_{0}) = 12 \cdot 1^{2} + 8 \cdot 1 = 20

u_{1} = 0.65

Δ u_{1} = ∣ 0.65 - 1 ∣ = 0.35

Δ u_{1} = 0.35

u_{2} = 0.47585 \dots : Δ u_{2} = 0.17414 \dots

f (u_{1}) = 4 \cdot 0.6 5^{3} + 4 \cdot 0.6 5^{2} - 1 = 1.7885

f^{^{'}} (u_{1}) = 12 \cdot 0.6 5^{2} + 8 \cdot 0.65 = 10.27

u_{2} = 0.47585 \dots

Δ u_{2} = ∣ 0.47585 \dots - 0.65 ∣ = 0.17414 \dots

Δ u_{2} = 0.17414 \dots

u_{3} = 0.42423 \dots : Δ u_{3} = 0.05161 \dots

f (u_{2}) = 4 \cdot 0.47585 \dots^{3} + 4 \cdot 0.47585 \dots^{2} - 1 = 0.33673 \dots

f^{^{'}} (u_{2}) = 12 \cdot 0.47585 \dots^{2} + 8 \cdot 0.47585 \dots = 6.52403 \dots

u_{3} = 0.42423 \dots

Δ u_{3} = ∣ 0.42423 \dots - 0.47585 \dots ∣ = 0.05161 \dots

Δ u_{3} = 0.05161 \dots

u_{4} = 0.41967 \dots : Δ u_{4} = 0.00455 \dots

f (u_{3}) = 4 \cdot 0.42423 \dots^{3} + 4 \cdot 0.42423 \dots^{2} - 1 = 0.02531 \dots

f^{^{'}} (u_{3}) = 12 \cdot 0.42423 \dots^{2} + 8 \cdot 0.42423 \dots = 5.55361 \dots

u_{4} = 0.41967 \dots

Δ u_{4} = ∣ 0.41967 \dots - 0.42423 \dots ∣ = 0.00455 \dots

Δ u_{4} = 0.00455 \dots

u_{5} = 0.41964 \dots : Δ u_{5} = 0.00003 \dots

f (u_{4}) = 4 \cdot 0.41967 \dots^{3} + 4 \cdot 0.41967 \dots^{2} - 1 = 0.00018 \dots

f^{^{'}} (u_{4}) = 12 \cdot 0.41967 \dots^{2} + 8 \cdot 0.41967 \dots = 5.47097 \dots

u_{5} = 0.41964 \dots

Δ u_{5} = ∣ 0.41964 \dots - 0.41967 \dots ∣ = 0.00003 \dots

Δ u_{5} = 0.00003 \dots

u_{6} = 0.41964 \dots : Δ u_{6} = 1.9624 E - 9

f (u_{5}) = 4 \cdot 0.41964 \dots^{3} + 4 \cdot 0.41964 \dots^{2} - 1 = 1.0735 E - 8

f^{^{'}} (u_{5}) = 12 \cdot 0.41964 \dots^{2} + 8 \cdot 0.41964 \dots = 5.47035 \dots

u_{6} = 0.41964 \dots

Δ u_{6} = ∣ 0.41964 \dots - 0.41964 \dots ∣ = 1.9624 E - 9

Δ u_{6} = 1.9624 E - 9

u \approx 0.41964 \dots

הפעל חילוק ארוך:

\frac{4 u ^{3} + 4 u ^{2} - 1}{u - 0.41964 \dots} = 4 u^{2} + 5.67857 \dots u + 2.38297 \dots

4 u^{2} + 5.67857 \dots u + 2.38297 \dots \approx 0

בשיטת ניטון-רפסון

4 u^{2} + 5.67857 \dots u + 2.38297 \dots = 0

מצא פתרון אחד ל:

u \in R

אין פתרון ל

4 u^{2} + 5.67857 \dots u + 2.38297 \dots = 0

הגדרת קירוב ניוטון-רפזון

f (u) = 4 u^{2} + 5.67857 \dots u + 2.38297 \dots

f^{^{'}} (u)

מצא את:

8 u + 5.67857 \dots

\frac{d}{d u} (4 u^{2} + 5.67857 \dots u + 2.38297 \dots)

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

:השתמש בחוק החיבור

= \frac{d}{d u} (4 u^{2}) + \frac{d}{d u} (5.67857 \dots u) + \frac{d}{d u} (2.38297 \dots)

\frac{d}{d u} (4 u^{2}) = 8 u

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

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

:הוצא את הקבוע

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

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

:השתמש בחוק החזקה

= 4 \cdot 2 u^{2 - 1}

פשט

= 8 u

\frac{d}{d u} (5.67857 \dots u) = 5.67857 \dots

\frac{d}{d u} (5.67857 \dots u)

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

:הוצא את הקבוע

= 5.67857 \dots \frac{d u}{d u}

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

:השתמש בנגזרת הבסיסית

= 5.67857 \dots \cdot 1

פשט

= 5.67857 \dots

\frac{d}{d u} (2.38297 \dots) = 0

\frac{d}{d u} (2.38297 \dots)

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

:נגזרת של קבוע

= 0

= 8 u + 5.67857 \dots + 0

פשט

= 8 u + 5.67857 \dots

u_{0} = 0

החלף

Δ u_{n + 1} < 0.000001

עד ש

u_{n + 1}

חשב

u_{1} = - 0.41964 \dots : Δ u_{1} = 0.41964 \dots

f (u_{0}) = 4 \cdot 0^{2} + 5.67857 \dots \cdot 0 + 2.38297 \dots = 2.38297 \dots

f^{^{'}} (u_{0}) = 8 \cdot 0 + 5.67857 \dots = 5.67857 \dots

u_{1} = - 0.41964 \dots

Δ u_{1} = ∣ - 0.41964 \dots - 0 ∣ = 0.41964 \dots

Δ u_{1} = 0.41964 \dots

u_{2} = - 0.72307 \dots : Δ u_{2} = 0.30343 \dots

f (u_{1}) = 4 (- 0.41964 \dots)^{2} + 5.67857 \dots (- 0.41964 \dots) + 2.38297 \dots = 0.70440 \dots

f^{^{'}} (u_{1}) = 8 (- 0.41964 \dots) + 5.67857 \dots = 2.32142 \dots

u_{2} = - 0.72307 \dots

Δ u_{2} = ∣ - 0.72307 \dots - (- 0.41964 \dots) ∣ = 0.30343 \dots

Δ u_{2} = 0.30343 \dots

u_{3} = 2.74959 \dots : Δ u_{3} = 3.47267 \dots

f (u_{2}) = 4 (- 0.72307 \dots)^{2} + 5.67857 \dots (- 0.72307 \dots) + 2.38297 \dots = 0.36829 \dots

f^{^{'}} (u_{2}) = 8 (- 0.72307 \dots) + 5.67857 \dots = - 0.10605 \dots

u_{3} = 2.74959 \dots

Δ u_{3} = ∣ 2.74959 \dots - (- 0.72307 \dots) ∣ = 3.47267 \dots

Δ u_{3} = 3.47267 \dots

u_{4} = 1.00660 \dots : Δ u_{4} = 1.74298 \dots

f (u_{3}) = 4 \cdot 2.74959 \dots^{2} + 5.67857 \dots \cdot 2.74959 \dots + 2.38297 \dots = 48.23776 \dots

f^{^{'}} (u_{3}) = 8 \cdot 2.74959 \dots + 5.67857 \dots = 27.67530 \dots

u_{4} = 1.00660 \dots

Δ u_{4} = ∣ 1.00660 \dots - 2.74959 \dots ∣ = 1.74298 \dots

Δ u_{4} = 1.74298 \dots

u_{5} = 0.12162 \dots : Δ u_{5} = 0.88498 \dots

f (u_{4}) = 4 \cdot 1.00660 \dots^{2} + 5.67857 \dots \cdot 1.00660 \dots + 2.38297 \dots = 12.15204 \dots

f^{^{'}} (u_{4}) = 8 \cdot 1.00660 \dots + 5.67857 \dots = 13.73139 \dots

u_{5} = 0.12162 \dots

Δ u_{5} = ∣ 0.12162 \dots - 1.00660 \dots ∣ = 0.88498 \dots

Δ u_{5} = 0.88498 \dots

u_{6} = - 0.34936 \dots : Δ u_{6} = 0.47098 \dots

f (u_{5}) = 4 \cdot 0.12162 \dots^{2} + 5.67857 \dots \cdot 0.12162 \dots + 2.38297 \dots = 3.13277 \dots

f^{^{'}} (u_{5}) = 8 \cdot 0.12162 \dots + 5.67857 \dots = 6.65153 \dots

u_{6} = - 0.34936 \dots

Δ u_{6} = ∣ - 0.34936 \dots - 0.12162 \dots ∣ = 0.47098 \dots

Δ u_{6} = 0.47098 \dots

u_{7} = - 0.65706 \dots : Δ u_{7} = 0.30770 \dots

f (u_{6}) = 4 (- 0.34936 \dots)^{2} + 5.67857 \dots (- 0.34936 \dots) + 2.38297 \dots = 0.88730 \dots

f^{^{'}} (u_{6}) = 8 (- 0.34936 \dots) + 5.67857 \dots = 2.88366 \dots

u_{7} = - 0.65706 \dots

Δ u_{7} = ∣ - 0.65706 \dots - (- 0.34936 \dots) ∣ = 0.30770 \dots

Δ u_{7} = 0.30770 \dots

u_{8} = - 1.55440 \dots : Δ u_{8} = 0.89734 \dots

f (u_{7}) = 4 (- 0.65706 \dots)^{2} + 5.67857 \dots (- 0.65706 \dots) + 2.38297 \dots = 0.37872 \dots

f^{^{'}} (u_{7}) = 8 (- 0.65706 \dots) + 5.67857 \dots = 0.42204 \dots

u_{8} = - 1.55440 \dots

Δ u_{8} = ∣ - 1.55440 \dots - (- 0.65706 \dots) ∣ = 0.89734 \dots

Δ u_{8} = 0.89734 \dots

u_{9} = - 1.07771 \dots : Δ u_{9} = 0.47669 \dots

f (u_{8}) = 4 (- 1.55440 \dots)^{2} + 5.67857 \dots (- 1.55440 \dots) + 2.38297 \dots = 3.22089 \dots

f^{^{'}} (u_{8}) = 8 (- 1.55440 \dots) + 5.67857 \dots = - 6.75669 \dots

u_{9} = - 1.07771 \dots

Δ u_{9} = ∣ - 1.07771 \dots - (- 1.55440 \dots) ∣ = 0.47669 \dots

Δ u_{9} = 0.47669 \dots

u_{10} = - 0.76886 \dots : Δ u_{10} = 0.30884 \dots

f (u_{9}) = 4 (- 1.07771 \dots)^{2} + 5.67857 \dots (- 1.07771 \dots) + 2.38297 \dots = 0.90896 \dots

f^{^{'}} (u_{9}) = 8 (- 1.07771 \dots) + 5.67857 \dots = - 2.94312 \dots

u_{10} = - 0.76886 \dots

Δ u_{10} = ∣ - 0.76886 \dots - (- 1.07771 \dots) ∣ = 0.30884 \dots

Δ u_{10} = 0.30884 \dots

u_{11} = 0.03881 \dots : Δ u_{11} = 0.80768 \dots

f (u_{10}) = 4 (- 0.76886 \dots)^{2} + 5.67857 \dots (- 0.76886 \dots) + 2.38297 \dots = 0.38153 \dots

f^{^{'}} (u_{10}) = 8 (- 0.76886 \dots) + 5.67857 \dots = - 0.47238 \dots

u_{11} = 0.03881 \dots

Δ u_{11} = ∣ 0.03881 \dots - (- 0.76886 \dots) ∣ = 0.80768 \dots

Δ u_{11} = 0.80768 \dots

u_{12} = - 0.39687 \dots : Δ u_{12} = 0.43569 \dots

f (u_{11}) = 4 \cdot 0.03881 \dots^{2} + 5.67857 \dots \cdot 0.03881 \dots + 2.38297 \dots = 2.60942 \dots

f^{^{'}} (u_{11}) = 8 \cdot 0.03881 \dots + 5.67857 \dots = 5.98911 \dots

u_{12} = - 0.39687 \dots

Δ u_{12} = ∣ - 0.39687 \dots - 0.03881 \dots ∣ = 0.43569 \dots

Δ u_{12} = 0.43569 \dots

u_{13} = - 0.70017 \dots : Δ u_{13} = 0.30329 \dots

f (u_{12}) = 4 (- 0.39687 \dots)^{2} + 5.67857 \dots (- 0.39687 \dots) + 2.38297 \dots = 0.75932 \dots

f^{^{'}} (u_{12}) = 8 (- 0.39687 \dots) + 5.67857 \dots = 2.50354 \dots

u_{13} = - 0.70017 \dots

Δ u_{13} = ∣ - 0.70017 \dots - (- 0.39687 \dots) ∣ = 0.30329 \dots

Δ u_{13} = 0.30329 \dots

u_{14} = - 5.46920 \dots : Δ u_{14} = 4.76902 \dots

f (u_{13}) = 4 (- 0.70017 \dots)^{2} + 5.67857 \dots (- 0.70017 \dots) + 2.38297 \dots = 0.36796 \dots

f^{^{'}} (u_{13}) = 8 (- 0.70017 \dots) + 5.67857 \dots = 0.07715 \dots

u_{14} = - 5.46920 \dots

Δ u_{14} = ∣ - 5.46920 \dots - (- 0.70017 \dots) ∣ = 4.76902 \dots

Δ u_{14} = 4.76902 \dots

לא יכול למצוא פתרון

The solutions are

u \approx 1.00000 \dots, u \approx 0.41964 \dots

u = cos (x)

החלף בחזרה

cos (x) \approx 1.00000 \dots, cos (x) \approx 0.41964 \dots

cos (x) \approx 1.00000 \dots, cos (x) \approx 0.41964 \dots

cos (x) = 1.00000 \dots : x = arccos (1.00000 \dots) + 2 πn, x = 2 π - arccos (1.00000 \dots) + 2 πn

cos (x) = 1.00000 \dots

Apply trig inverse properties

cos (x) = 1.00000 \dots

cos (x) = 1.00000 \dots :

פתרונות כלליים עבור

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (1.00000 \dots) + 2 πn, x = 2 π - arccos (1.00000 \dots) + 2 πn

x = arccos (1.00000 \dots) + 2 πn, x = 2 π - arccos (1.00000 \dots) + 2 πn

cos (x) = 0.41964 \dots : x = arccos (0.41964 \dots) + 2 πn, x = 2 π - arccos (0.41964 \dots) + 2 πn

cos (x) = 0.41964 \dots

Apply trig inverse properties

cos (x) = 0.41964 \dots

cos (x) = 0.41964 \dots :

פתרונות כלליים עבור

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.41964 \dots) + 2 πn, x = 2 π - arccos (0.41964 \dots) + 2 πn

x = arccos (0.41964 \dots) + 2 πn, x = 2 π - arccos (0.41964 \dots) + 2 πn

אחד את הפתרונות

x = arccos (1.00000 \dots) + 2 πn, x = 2 π - arccos (1.00000 \dots) + 2 πn, x = arccos (0.41964 \dots) + 2 πn, x = 2 π - arccos (0.41964 \dots) + 2 πn

הראה פיתרון ביצוג עשרוני

x = 0 + 2 πn, x = 2 π - 0 + 2 πn, x = 1.13774 \dots + 2 πn, x = 2 π - 1.13774 \dots + 2 πn

דוגמאות פופולריות

sin(θ)=sin(-(2pi)/3)

sin (θ) = sin (- \frac{2 π}{3})

sin(x)=(23/24)

sin (x) = (\frac{23}{24})

tan(θ)=-5/4

tan (θ) = - \frac{5}{4}

5-6sin^2(x)-cos(x)=0

5 - 6 sin^{2} (x) - cos (x) = 0

cos(x)+tan(x)sin(x)=2

cos (x) + tan (x) sin (x) = 2