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شائع علم المثلثات >

tan(x)*sin(x)> 1/(2cos(x))

الحلّ

tan (x) \cdot sin (x) > \frac{1}{2 cos ( x )}

الحلّ

\frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{7 π}{4} + 2 πn

تدوين الفاصل الزمني

(\frac{π}{4} + 2 πn, \frac{π}{2} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{5 π}{4} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, \frac{7 π}{4} + 2 πn)

عشري

0.78539 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.92699 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 5.49778 \dots + 2 πn

خطوات الحلّ

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

انقل

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

إلى الجانب الأيسر

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

من الطرفين

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

اطرح

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

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

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

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

دوريّة:

2 π

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

tan (x) sin (x), \frac{1}{2 cos ( x )}

tan (x) sin (x)

دوريّة:

2 π

:

مركّبة من الدوالّ والدوريّات التالية

tan (x) sin (x)

π

مع دوريّات

tan (x)

:

الدوريّة المركّبة للدالّة هي

2 π

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

دوريّة:

2 π

a cos (b x \pm c) \pm d = \frac{cos دوريّة}{∣ b ∣}

دوريّة ال

2 π

هي

cos (x)

دوريّة

= \frac{2 π}{∣ 1 ∣}

بسّط

= 2 π

Combine periods:

2 π, 2 π

= 2 π

sin, cos :

عبّر بواسطة

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

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

:Use the basic trigonometric identity

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

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

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

بسّط:

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

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

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

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

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

:اضرب كسور

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

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

sin (x) sin (x)

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

:فعّل قانون القوى

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

= sin^{1 + 1} (x)

1 + 1 = 2 :

اجمع الأعداد

= sin^{2} (x)

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

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

cos (x), 2 cos (x)

المضاعف المشترك الأصغر لـ:

2 cos (x)

cos (x), 2 cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

2 cos (x)

= 2 cos (x)

اكتب مجددًا الكسور بحيث يكون المقام مشترك

2 cos (x)

اضرب كل بسط ومقام بتعبير الذي يؤدّي إلى مقام مشترك

For

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

multiply the denominator and numerator by

2

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

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{2 sin ^{2} ( x ) - 1}{2 cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

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

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

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

بالاستعانة بطريقة التعويض

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

sin (x) = u :

على افتراض أنّ

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

انقل

1

إلى الجانب الأيمن

2 u^{2} - 1 = 0

للطرفين

1

أضف

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

بسّط

2 u^{2} = 1

2 u^{2} = 1

2

اقسم الطرفين على

2 u^{2} = 1

2

اقسم الطرفين على

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

بسّط

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

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

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

الحلول هي

x^{2} = f (a)

لـ

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 < 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x < 2 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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

0 \leq x < 2 π :

حلول للمدى

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = \frac{5 π}{4}, x = \frac{7 π}{4}

وحّد الحلول

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

2 cos (x) = 0

2

اقسم الطرفين على

2 cos (x) = 0

2

اقسم الطرفين على

\frac{2 cos ( x )}{2} = \frac{0}{2}

بسّط

cos (x) = 0

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}

\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}

ميّز المقاطع المختلفة

0 < 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}{4}, \frac{5 π}{4} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < \frac{7 π}{4}, \frac{7 π}{4} < x < 2 π

لخّص في جدول

2 sin^{2} (x) - 1 cos (x) \frac{2 s i n ^{2} ( x ) - 1}{2 c o s ( x )} x = 0 - + - 0 < 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}{4} - - + x = \frac{5 π}{4} 0 - 0 \frac{5 π}{4} < x < \frac{3 π}{2} + - - x = \frac{3 π}{2} + 0 غير معرّف \frac{3 π}{2} < x < \frac{7 π}{4} + + + x = \frac{7 π}{4} 0 + 0 \frac{7 π}{4} < x < 2 π - + - x = 2 π - + -

> 0 :

ميّز المقاطع التي تحقّق الشرط

\frac{π}{4} < x < \frac{π}{2} or \frac{3 π}{4} < x < \frac{5 π}{4} or \frac{3 π}{2} < x < \frac{7 π}{4}

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

استخدم دوريّة الـ

\frac{π}{4} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{7 π}{4} + 2 πn

أمثلة شائعة

solvefor c,sin(xcos(2x))<= 1

so l v e f or c, sin (x cos (2 x)) \leq 1

cos(5x)cos(x/4)-sin(5x)sin(x/4)>=-(sqrt(2))/2

cos (5 x) cos (\frac{x}{4}) - sin (5 x) sin (\frac{x}{4}) \geq - \frac{2}{2}

2sin(x/2)-1>= 0

2 sin (\frac{x}{2}) - 1 \geq 0

sin(2x-(3pi)/2)<= 0

sin (2 x - \frac{3 π}{2}) \leq 0

tan(-3x)<1

tan (- 3 x) < 1