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شائع علم المثلثات >

(4cos^2(x)-3)(1-tan^2(x))<= 0

الحلّ

(4 cos^{2} (x) - 3) (1 - tan^{2} (x)) \leq 0

الحلّ

\frac{π}{6} + πn \leq x \leq \frac{π}{4} + πn or \frac{3 π}{4} + πn \leq x \leq \frac{5 π}{6} + πn

تدوين الفاصل الزمني

[\frac{π}{6} + πn, \frac{π}{4} + πn] \cup [\frac{3 π}{4} + πn, \frac{5 π}{6} + πn]

عشري

0.52359 \dots + πn \leq x \leq 0.78539 \dots + πn or 2.35619 \dots + πn \leq x \leq 2.61799 \dots + πn

خطوات الحلّ

(4 cos^{2} (x) - 3) (1 - tan^{2} (x)) \leq 0

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

:استخدم المتطابقة التالية

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

لذلك

(4 (1 - sin^{2} (x)) - 3) (1 - tan^{2} (x)) \leq 0

(4 (1 - sin^{2} (x)) - 3) (1 - tan^{2} (x))

بسّط:

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

(4 (1 - sin^{2} (x)) - 3) (1 - tan^{2} (x))

4 (1 - sin^{2} (x)) - 3

وسٌع:

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

4 (1 - sin^{2} (x)) - 3

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) - 3

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

بسّط:

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

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

جمّع التعابير المتشابهة

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

4 - 3 = 1 :

اطرح/اجمع الأعداد

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

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

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

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

(- 4 sin^{2} (x) + 1) (1 - tan^{2} (x)) \leq 0

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

دوريّة:

π

:

مركّبة من الدوالّ والدوريّات التالية

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

2 π

مع دوريّات

sin (x)

:

الدوريّة المركّبة للدالّة هي

= π

sin, cos :

عبّر بواسطة

(- 4 sin^{2} (x) + 1) (1 - tan^{2} (x)) \leq 0

tan (x) = \frac{sin ( x )}{cos ( x )}

:Use the basic trigonometric identity

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

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

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

بسّط:

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

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

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

:فعّل قانون القوى

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

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

وحّد:

\frac{cos ^{2} ( x ) - sin ^{2} ( x )}{cos ^{2} ( x )}

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

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos^{2} (x) = cos^{2} (x) :

اضرب

= \frac{cos ^{2} ( x ) - sin ^{2} ( x )}{cos ^{2} ( x )}

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

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

:اضرب كسور

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

\frac{( cos ^{2} ( x ) - sin ^{2} ( x ) ) ( - 4 sin ^{2} ( x ) + 1 )}{cos ^{2} ( x )} = 0, 0 \leq x < π : x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{π}{6}, x = \frac{5 π}{6}

\frac{( cos ^{2} ( x ) - sin ^{2} ( x ) ) ( - 4 sin ^{2} ( x ) + 1 )}{cos ^{2} ( x )} = 0, 0 \leq x < π

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

(cos^{2} (x) - sin^{2} (x)) (- 4 sin^{2} (x) + 1) = 0

حلّ كل جزء على حدة

cos^{2} (x) - sin^{2} (x) = 0 or - 4 sin^{2} (x) + 1 = 0

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

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

Rewrite using trig identities

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

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

:فعّل متطابقة الزاوية المضاعفة

= cos (2 x)

cos (2 x) = 0

cos (2 x) = 0 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

2 x = \frac{π}{2} + 2 πn

حلّ:

x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

2

اقسم الطرفين على

2 x = \frac{π}{2} + 2 πn

2

اقسم الطرفين على

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

بسّط

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

\frac{2 x}{2}

بسّط:

x

\frac{2 x}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= x

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

بسّط:

\frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

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

: استخدم ميزات الكسور التالية

= \frac{π}{2 \cdot 2}

2 \cdot 2 = 4 :

اضرب الأعداد

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

حلّ:

x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

2

اقسم الطرفين على

2 x = \frac{3 π}{2} + 2 πn

2

اقسم الطرفين على

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

بسّط

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

\frac{2 x}{2}

بسّط:

x

\frac{2 x}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= x

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

بسّط:

\frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

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

\frac{\frac{3 π}{2}}{2}

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

: استخدم ميزات الكسور التالية

= \frac{3 π}{2 \cdot 2}

2 \cdot 2 = 4 :

اضرب الأعداد

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

\frac{2}{2} = 1 :

اقسم الأعداد

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

0 \leq x < π :

حلول للمدى

x = \frac{π}{4}, x = \frac{3 π}{4}

- 4 sin^{2} (x) + 1 = 0, 0 \leq x < π : x = \frac{π}{6}, x = \frac{5 π}{6}

- 4 sin^{2} (x) + 1 = 0, 0 \leq x < π

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

- 4 u^{2} + 1 = 0

- 4 u^{2} + 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

- 4 u^{2} + 1 = 0

انقل

1

إلى الجانب الأيمن

- 4 u^{2} + 1 = 0

من الطرفين

1

اطرح

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

بسّط

- 4 u^{2} = - 1

- 4 u^{2} = - 1

- 4

اقسم الطرفين على

- 4 u^{2} = - 1

- 4

اقسم الطرفين على

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

بسّط

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

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

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

= \frac{1}{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{1}{2}

1 = 1

فعّل القانون

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

\frac{1}{4}

بسّط:

\frac{1}{2}

\frac{1}{4}

a \geq 0, b \geq 0

بافتراض أنّ

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

:فعّل قانون الجذور

= \frac{1}{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{1}{2}

= - \frac{1}{2}

1 = 1

فعّل القانون

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, 0 \leq x < π : x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = \frac{1}{2}, 0 \leq x < π

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

0 \leq x < π :

حلول للمدى

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x < π :

لا يوجد حلّ

sin (x) = - \frac{1}{2}, 0 \leq x < π

sin (x) = - \frac{1}{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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

0 \leq x < π :

حلول للمدى

لا يوجد حلّ

وحّد الحلول

x = \frac{π}{6}, x = \frac{5 π}{6}

وحّد الحلول

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{π}{6}, x = \frac{5 π}{6}

Find the undefined points:

x = \frac{π}{2}

Find the zeros of the denominator

cos^{2} (x) = 0

x^{n} = 0 \Rightarrow x = 0

فعّل القانون

cos (x) = 0

cos (x) = 0 :

حلول عامّة لـ

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

0 \leq x < π :

حلول للمدى

x = \frac{π}{2}

\frac{π}{6}, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{6}

ميّز المقاطع المختلفة

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < π

لخّص في جدول

cos^{2} (x) - sin^{2} (x) - 4 sin^{2} (x) + 1 cos^{2} (x) \frac{( c o s ^{2} ( x ) - s i n ^{2} ( x ) ) ( - 4 s i n ^{2} ( x ) + 1 )}{c o s ^{2} ( x )} x = 0 + + + + 0 < x < \frac{π}{6} + + + + x = \frac{π}{6} + 0 + 0 \frac{π}{6} < x < \frac{π}{4} + - + - x = \frac{π}{4} 0 - + 0 \frac{π}{4} < x < \frac{π}{2} - - + + x = \frac{π}{2} - - 0 غير معرّف \frac{π}{2} < x < \frac{3 π}{4} - - + + x = \frac{3 π}{4} 0 - + 0 \frac{3 π}{4} < x < \frac{5 π}{6} + - + - x = \frac{5 π}{6} + 0 + 0 \frac{5 π}{6} < x < π + + + + x = π + + + +

\leq 0 :

ميّز المقاطع التي تحقّق الشرط

x = \frac{π}{6} or \frac{π}{6} < x < \frac{π}{4} or x = \frac{π}{4} or x = \frac{3 π}{4} or \frac{3 π}{4} < x < \frac{5 π}{6} or x = \frac{5 π}{6}

ادمج المجالات المتطابقة

\frac{π}{6} \leq x \leq \frac{π}{4} or \frac{3 π}{4} \leq x < \frac{5 π}{6} or x = \frac{5 π}{6}