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شائع علم المثلثات >

tan(x)<= cos(x)

الحلّ

tan (x) \leq cos (x)

الحلّ

2 πn \leq x \leq 0.66623 \dots + 2 πn or \frac{π}{2} + 2 πn < x \leq π - 0.66623 \dots + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

تدوين الفاصل الزمني

[2 πn, 0.66623 \dots + 2 πn] \cup (\frac{π}{2} + 2 πn, π - 0.66623 \dots + 2 πn] \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn]

عشري

2 πn \leq x \leq 0.66623 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.47535 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

خطوات الحلّ

tan (x) \leq cos (x)

انقل

cos (x)

إلى الجانب الأيسر

tan (x) \leq cos (x)

من الطرفين

cos (x)

اطرح

tan (x) - cos (x) \leq cos (x) - cos (x)

tan (x) - cos (x) \leq 0

tan (x) - cos (x) \leq 0

tan (x) - cos (x)

دوريّة:

2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

tan (x), cos (x)

tan (x)

دوريّة:

π

π

هي

tan (x)

دوريّة

= π

cos (x)

دوريّة:

2 π

2 π

هي

cos (x)

دوريّة

= 2 π

Combine periods:

π, 2 π

= 2 π

sin, cos :

عبّر بواسطة

tan (x) - cos (x) \leq 0

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

:Use the basic trigonometric identity

\frac{sin ( x )}{cos ( x )} - cos (x) \leq 0

\frac{sin ( x )}{cos ( x )} - cos (x) \leq 0

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

بسّط:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

sin (x) - cos (x) cos (x)

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

cos (x) cos (x)

a^{b} \cdot a^{c} = a^{b + c}

:فعّل قانون القوى

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

= cos^{1 + 1} (x)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (x)

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{sin ( x ) - cos ^{2} ( x )}{cos ( x )} = 0, 0 \leq x < 2 π : x = 0.66623 \dots, x = π - 0.66623 \dots

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

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

افتح أقواس

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

فعّل قوانين سالب-موجب

- (- a) = a, - (a) = - a

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

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

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

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

a x^{2} + b x + c = 0

اكتب بالصورة الاعتياديّة

u^{2} + u - 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

u^{2} + u - 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 1, b = 1, c = - 1

لـ

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

- (- a) = a

فعّل القانون

= 1 + 4 \cdot 1 \cdot 1

4 \cdot 1 \cdot 1 = 4 :

اضرب الأعداد

= 1 + 4

1 + 4 = 5 :

اجمع الأعداد

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

2 \cdot 1 = 2 :

اضرب الأعداد

= \frac{- 1 - 5}{2}

حلول المعادلة التربيعيّة هي

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

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

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

Apply trig inverse properties

sin (x) = \frac{- 1 + 5}{2}

sin (x) = \frac{- 1 + 5}{2} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

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

لا يوجد حلّ

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

- 1 \leq sin (x) \leq 1

لا يوجد حلّ

وحّد الحلول

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

أظهر الحلّ بالتمثيل العشريّ

x = 0.66623 \dots, x = π - 0.66623 \dots

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

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 < 2 π :

حلول للمدى

x = \frac{π}{2}, x = \frac{3 π}{2}

0.66623 \dots, \frac{π}{2}, π - 0.66623 \dots, \frac{3 π}{2}

ميّز المقاطع المختلفة

0 < x < 0.66623 \dots, 0.66623 \dots < x < \frac{π}{2}, \frac{π}{2} < x < π - 0.66623 \dots, π - 0.66623 \dots < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

لخّص في جدول

sin (x) - cos^{2} (x) cos (x) \frac{s i n ( x ) - c o s ^{2} ( x )}{c o s ( x )} x = 0 - + - 0 < x < 0.66623 \dots - + - x = 0.66623 \dots 0 + 0 0.66623 \dots < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 غير معرّف \frac{π}{2} < x < π - 0.66623 \dots + - - x = π - 0.66623 \dots 0 - 0 π - 0.66623 \dots < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} - 0 غير معرّف \frac{3 π}{2} < x < 2 π - + - x = 2 π - + -

\leq 0 :

ميّز المقاطع التي تحقّق الشرط

x = 0 or 0 < x < 0.66623 \dots or x = 0.66623 \dots or \frac{π}{2} < x < π - 0.66623 \dots or x = π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π or x = 2 π

ادمج المجالات المتطابقة

0 \leq x \leq 0.66623 \dots or \frac{π}{2} < x \leq π - 0.66623 \dots or \frac{3 π}{2} < x < 2 π or x = 2 π