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شائع علم المثلثات >

(sin(2x)(4cos^2(x)-1))/(sin(x))>0

الحلّ

\frac{sin ( 2 x ) ( 4 cos ^{2} ( x ) - 1 )}{sin ( x )} > 0

الحلّ

2 πn < x < \frac{π}{3} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

تدوين الفاصل الزمني

(2 πn, \frac{π}{3} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{2 π}{3} + 2 πn) \cup (\frac{4 π}{3} + 2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{5 π}{3} + 2 πn, 2 π + 2 πn)

عشري

2 πn < x < 1.04719 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 2.09439 \dots + 2 πn or 4.18879 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 5.23598 \dots + 2 πn < x < 6.28318 \dots + 2 πn

خطوات الحلّ

\frac{sin ( 2 x ) ( 4 cos ^{2} ( x ) - 1 )}{sin ( x )} > 0

cos^{2} (x) + sin^{2} (x) = 1

:استخدم المتطابقة التالية

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

لذلك

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )} > 0

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )}

بسّط:

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

\frac{sin ( 2 x ) ( 4 ( 1 - sin ^{2} ( x ) ) - 1 )}{sin ( x )}

4 (1 - sin^{2} (x)) - 1

وسٌع:

- 4 sin^{2} (x) + 3

4 (1 - sin^{2} (x)) - 1

4 (1 - sin^{2} (x))

وسٌع:

4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

4 \cdot 1 = 4 :

اضرب الأعداد

= 4 - 4 sin^{2} (x)

= 4 - 4 sin^{2} (x) - 1

4 - 4 sin^{2} (x) - 1

بسّط:

- 4 sin^{2} (x) + 3

4 - 4 sin^{2} (x) - 1

جمّع التعابير المتشابهة

= - 4 sin^{2} (x) + 4 - 1

4 - 1 = 3 :

اطرح/اجمع الأعداد

= - 4 sin^{2} (x) + 3

= - 4 sin^{2} (x) + 3

= \frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} > 0

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

دوريّة:

2 π

:

مركّبة من الدوالّ والدوريّات التالية

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

\frac{2 π}{2}

مع دوريّات

sin (2 x)

:

الدوريّة المركّبة للدالّة هي

= 2 π

Find the zeroes and undifined points of

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 x) (- 4 sin^{2} (x) + 3) = 0

حلّ كل جزء على حدة

sin (2 x) = 0 or - 4 sin^{2} (x) + 3 = 0

sin (2 x) = 0, 0 \leq x < 2 π : x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}

sin (2 x) = 0, 0 \leq x < 2 π

sin (2 x) = 0 :

حلول عامّة لـ

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn

حلّ:

x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

2

اقسم الطرفين على

2 x = 2 πn

2

اقسم الطرفين على

\frac{2 x}{2} = \frac{2 πn}{2}

بسّط

x = πn

x = πn

2 x = π + 2 πn

حلّ:

x = \frac{π}{2} + πn

2 x = π + 2 πn

2

اقسم الطرفين على

2 x = π + 2 πn

2

اقسم الطرفين على

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

بسّط

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

0 \leq x < 2 π :

حلول للمدى

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}

- 4 sin^{2} (x) + 3 = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

- 4 sin^{2} (x) + 3 = 0, 0 \leq x < 2 π

بالاستعانة بطريقة التعويض

- 4 sin^{2} (x) + 3 = 0

sin (x) = u :

على افتراض أنّ

- 4 u^{2} + 3 = 0

- 4 u^{2} + 3 = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 4 u^{2} + 3 = 0

انقل

3

إلى الجانب الأيمن

- 4 u^{2} + 3 = 0

من الطرفين

3

اطرح

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

بسّط

- 4 u^{2} = - 3

- 4 u^{2} = - 3

- 4

اقسم الطرفين على

- 4 u^{2} = - 3

- 4

اقسم الطرفين على

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

بسّط

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

x = f (a), - f (a)

الحلول هي

x^{2} = f (a)

لـ

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{3}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

\frac{3}{4}

بسّط:

\frac{3}{2}

\frac{3}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{3}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{2 π}{3}

sin (x) = \frac{3}{2}, 0 \leq x < 2 π

sin (x) = \frac{3}{2} :

حلول عامّة لـ

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = \frac{π}{3}, x = \frac{2 π}{3}

sin (x) = - \frac{3}{2}, 0 \leq x < 2 π : x = \frac{4 π}{3}, x = \frac{5 π}{3}

sin (x) = - \frac{3}{2}, 0 \leq x < 2 π

sin (x) = - \frac{3}{2} :

حلول عامّة لـ

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = \frac{4 π}{3}, x = \frac{5 π}{3}

وحّد الحلول

x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

وحّد الحلول

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

0, π :

بما أنّ المعادلة غير معرّفة لـ

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Find the undefined points:

x = 0, x = π

Find the zeros of the denominator

sin (x) = 0

sin (x) = 0 :

حلول عامّة لـ

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn

حلّ:

x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = 0, x = π

0, \frac{π}{3}, \frac{π}{2}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{3 π}{2}, \frac{5 π}{3}

ميّز المقاطع المختلفة

0 < x < \frac{π}{3}, \frac{π}{3} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{2 π}{3}, \frac{2 π}{3} < x < π, π < x < \frac{4 π}{3}, \frac{4 π}{3} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < \frac{5 π}{3}, \frac{5 π}{3} < x < 2 π

لخّص في جدول

sin (2 x) - 4 sin^{2} (x) + 3 sin (x) \frac{s i n ( 2 x ) ( - 4 s i n ^{2} ( x ) + 3 )}{s i n ( x )} x = 0 0 + 0 غير معرّف 0 < x < \frac{π}{3} + + + + x = \frac{π}{3} + 0 + 0 \frac{π}{3} < x < \frac{π}{2} + - + - x = \frac{π}{2} 0 - + 0 \frac{π}{2} < x < \frac{2 π}{3} - - + + x = \frac{2 π}{3} - 0 + 0 \frac{2 π}{3} < x < π - + + - x = π 0 + 0 غير معرّف π < x < \frac{4 π}{3} + + - - x = \frac{4 π}{3} + 0 - 0 \frac{4 π}{3} < x < \frac{3 π}{2} + - - + x = \frac{3 π}{2} 0 - - 0 \frac{3 π}{2} < x < \frac{5 π}{3} - - - - x = \frac{5 π}{3} - 0 - 0 \frac{5 π}{3} < x < 2 π - + - + x = 2 π 0 + 0 غير معرّف

> 0 :

ميّز المقاطع التي تحقّق الشرط

0 < x < \frac{π}{3} or \frac{π}{2} < x < \frac{2 π}{3} or \frac{4 π}{3} < x < \frac{3 π}{2} or \frac{5 π}{3} < x < 2 π

\frac{sin ( 2 x ) ( - 4 sin ^{2} ( x ) + 3 )}{sin ( x )} :

استخدم دوريّة الـ

2 πn < x < \frac{π}{3} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

أمثلة شائعة

cot(x)>cot(1/x)

cot (x) > cot (\frac{1}{x})

(2sin(x)+sqrt(2))/(cos(x))<= 0

\frac{2 sin ( x ) + 2}{cos ( x )} \leq 0

1/2 >cos(x)

\frac{1}{2} > cos (x)

2cos^2(a)-1>=-1/8

2 cos^{2} (a) - 1 \geq - \frac{1}{8}

4sin^2(x)-3<0,-2pi<= x<= 0

4 sin^{2} (x) - 3 < 0, - 2 π \leq x \leq 0