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עברית

العربية

(cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1

الحلّ

\frac{cos ^{2022} ( x )}{( sin ^{2022} ( x ) + cos ^{2022} ( x ) )} = 1

x = 2 πn, x = π + 2 πn

درجات

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

\frac{cos ^{2022} ( x )}{( sin ^{2022} ( x ) + cos ^{2022} ( x ) )} = 1

من الطرفين

1

اطرح

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1 = 0

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1

بسّط:

- \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1

1 = \frac{1 ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

:حوّل الأعداد لكسور

= \frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - \frac{1 \cdot ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{cos ^{2022} ( x ) - 1 \cdot ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

1 \cdot (sin^{2022} (x) + cos^{2022} (x)) = (sin^{2022} (x) + cos^{2022} (x)) :

اضرب

= \frac{cos ^{2022} ( x ) - ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

cos^{2022} (x) - (sin^{2022} (x) + cos^{2022} (x))

وسٌع:

- sin^{2022} (x)

cos^{2022} (x) - (sin^{2022} (x) + cos^{2022} (x))

- (sin^{2022} (x) + cos^{2022} (x)) : - sin^{2022} (x) - cos^{2022} (x)

- (sin^{2022} (x) + cos^{2022} (x))

افتح أقواس

= - (sin^{2022} (x)) - (cos^{2022} (x))

فعّل قوانين سالب-موجب

+ (- a) = - a

= - sin^{2022} (x) - cos^{2022} (x)

= cos^{2022} (x) - sin^{2022} (x) - cos^{2022} (x)

cos^{2022} (x) - cos^{2022} (x) = 0 :

اجمع العناصر المتشابهة

= - sin^{2022} (x)

= \frac{- sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

\frac{- a}{b} = - \frac{a}{b}

: استخدم ميزات الكسور التالية

= - \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

- \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} = 0

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

- (sin^{2022} (x)) = 0

- 1

اقسم الطرفين على

- (sin^{2022} (x)) = 0

- 1

اقسم الطرفين على

\frac{- sin ^{2022} ( x )}{- 1} = \frac{0}{- 1}

بسّط

sin^{2022} (x) = 0

sin^{2022} (x) = 0

حلّ عن طريق مساواة العوامل لصفر

sin (x) = 0

الحل للمعادلة هو

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