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شائع علم المثلثات >

1-sin^2(x)-cos(2x)= 1/2

الحلّ

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

الحلّ

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

درجات

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

خطوات الحلّ

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

من الطرفين

\frac{1}{2}

اطرح

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

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

بسّط:

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

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

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

Rewrite using trig identities

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

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

:فعّل متطابقة الزاوية المضاعفة

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

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

1 - 2 cos (2 x) + 2 (cos (2 x) - cos^{2} (x))

بسّط:

1 - 2 cos^{2} (x)

1 - 2 cos (2 x) + 2 (cos (2 x) - cos^{2} (x))

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

وسٌع:

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

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

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = 2, b = cos (2 x), c = cos^{2} (x)

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

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

- 2 cos (2 x) + 2 cos (2 x) = 0 :

اجمع العناصر المتشابهة

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

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

1 - 2 u^{2} = 0

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

1 - 2 u^{2} = 0

انقل

1

إلى الجانب الأيمن

1 - 2 u^{2} = 0

من الطرفين

1

اطرح

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

استبدل مجددًا

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

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

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

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

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

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

حلول عامّة لـ

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

وحّد الحلول

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

أظهر الحلّ بالتمثيل العشريّ

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

رسم

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