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csc(θ)=-5/4 \land cos(θ)>0

الحلّ

csc (θ) = - \frac{5}{4} and cos (θ) > 0

الحلّ

θ = - 0.92729 \dots + 2 πn

+1

تدوين الفاصل الزمني

θ = - 0.92729 \dots + 2 πn

خطوات الحلّ

csc (θ) = - \frac{5}{4} and cos (θ) > 0

csc (θ) = - \frac{5}{4} : θ = - 0.92729 \dots + 2 πn, θ = π + 0.92729 \dots + 2 πn

csc (θ) = - \frac{5}{4}

Apply trig inverse properties

csc (θ) = - \frac{5}{4}

csc (θ) = - \frac{5}{4} :

حلول عامّة لـ

csc (x) = - a \Rightarrow x = arccsc (- a) + 2 πn, x = π + arccsc (a) + 2 πn

θ = arccsc (- \frac{5}{4}) + 2 πn, θ = π + arccsc (\frac{5}{4}) + 2 πn

θ = arccsc (- \frac{5}{4}) + 2 πn, θ = π + arccsc (\frac{5}{4}) + 2 πn

أظهر الحلّ بالتمثيل العشريّ

θ = - 0.92729 \dots + 2 πn, θ = π + 0.92729 \dots + 2 πn

cos (θ) > 0 : - \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

cos (θ) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < θ < arccos (0) + 2 πn

- arccos (0)

بسّط:

- \frac{π}{2}

- arccos (0)

استخدم المتطابقة الأساسيّة التالية:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

arccos (0)

بسّط:

\frac{π}{2}

arccos (0)

استخدم المتطابقة الأساسيّة التالية:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

- \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

وحّد المقاطع

(θ = π + 0.92729 \dots + 2 πn or θ = - 0.92729 \dots + 2 πn) and - \frac{π}{2} + 2 πn < θ < \frac{π}{2} + 2 πn

ادمج المجالات المتطابقة

θ = - 0.92729 \dots + 2 πn

أمثلة شائعة

sin(θ)<0\land tan(θ)<0

sin (θ) < 0 and tan (θ) < 0

sin (t) 0 \leq t < π

csc(θ)=4\land cot(θ)<0

csc (θ) = 4 and cot (θ) < 0

-(sqrt(2))/2 <sin(x/2)<(sqrt(2))/2

- \frac{2}{2} < sin (\frac{x}{2}) < \frac{2}{2}

-sqrt(3)<= tan(x)<= ((sqrt(3)))/3

- 3 \leq tan (x) \leq \frac{( 3 )}{3}