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شائع علم المثلثات >

solvefor x,d^2+3d+2y=4cos^2(x)

الحلّ

solve for

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

الحلّ

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

خطوات الحلّ

d^{2} + 3 d + 2 y = 4 cos^{2} (x)

بدّل الأطراف

4 cos^{2} (x) = d^{2} + 3 d + 2 y

بالاستعانة بطريقة التعويض

4 cos^{2} (x) = d^{2} + 3 d + 2 y

cos (x) = u :

على افتراض أنّ

4 u^{2} = d^{2} + 3 d + 2 y

4 u^{2} = d^{2} + 3 d + 2 y : u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

4 u^{2} = d^{2} + 3 d + 2 y

4

اقسم الطرفين على

4 u^{2} = d^{2} + 3 d + 2 y

4

اقسم الطرفين على

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

بسّط

\frac{4 u ^{2}}{4} = \frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

\frac{4 u ^{2}}{4}

بسّط:

u^{2}

\frac{4 u ^{2}}{4}

\frac{4}{4} = 1 :

اقسم الأعداد

= u^{2}

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

بسّط:

\frac{d ^{2} + 3 d + 2 y}{4}

\frac{d ^{2}}{4} + \frac{3 d}{4} + \frac{2 y}{4}

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

فعّل القانون

= \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

u^{2} = \frac{d ^{2} + 3 d + 2 y}{4}

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

الحلول هي

x^{2} = f (a)

لـ

u = \frac{d ^{2} + 3 d + 2 y}{4}, u = - \frac{d ^{2} + 3 d + 2 y}{4}

\frac{d ^{2} + 3 d + 2 y}{4}

بسّط:

\frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{d ^{2} + 3 d + 2 y}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{d ^{2} + 3 d + 2 y}{2}

- \frac{d ^{2} + 3 d + 2 y}{4}

بسّط:

- \frac{d ^{2} + 3 d + 2 y}{2}

- \frac{d ^{2} + 3 d + 2 y}{4}

\frac{d ^{2} + 3 d + 2 y}{4}

بسّط:

\frac{d ^{2} + 3 d + 2 y}{2}

\frac{d ^{2} + 3 d + 2 y}{4}

a \geq 0, b \geq 0

بافتراض أنّ

n \frac{a}{b} = \frac{n a}{n b}

:فعّل قانون الجذور

= \frac{d ^{2} + 3 d + 2 y}{4}

4 = 2

4

4 = 2^{2} :

حلّل العدد لعوامله أوّليّة

= 2^{2}

n a^{n} = a

:فعْل قانون الجذور

2^{2} = 2

= 2

= \frac{d ^{2} + 3 d + 2 y}{2}

= - \frac{2 y + d ^{2} + 3 d}{2}

= - \frac{d ^{2} + 3 d + 2 y}{2}

u = \frac{d ^{2} + 3 d + 2 y}{2}, u = - \frac{d ^{2} + 3 d + 2 y}{2}

u = cos (x)

استبدل مجددًا

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}, cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

Apply trig inverse properties

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = \frac{d ^{2} + 3 d + 2 y}{2} :

حلول عامّة لـ

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

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2} : x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

Apply trig inverse properties

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2}

cos (x) = - \frac{d ^{2} + 3 d + 2 y}{2} :

حلول عامّة لـ

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

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

وحّد الحلول

x = arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (\frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn, x = - arccos (- \frac{d ^{2} + 3 d + 2 y}{2}) + 2 πn

رسم

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