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شائع >

solvefor t,f=3cos(pttheta)-2

الحلّ

solve for

t, f = 3 cos (ptt h e t a) - 2

الحلّ

t = 3 \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}, t = 3 \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}

خطوات الحلّ

f = 3 cos (ptt h e t a) - 2

بدّل الأطراف

3 cos (ptt h e t a) - 2 = f

انقل

2

إلى الجانب الأيمن

3 cos (ptt h e t a) - 2 = f

للطرفين

2

أضف

3 cos (ptt h e t a) - 2 + 2 = f + 2

بسّط

3 cos (ptt h e t a) = f + 2

3 cos (ptt h e t a) = f + 2

3

اقسم الطرفين على

3 cos (ptt h e t a) = f + 2

3

اقسم الطرفين على

\frac{3 cos ( ptt h e t a )}{3} = \frac{f}{3} + \frac{2}{3}

بسّط

\frac{3 cos ( ptt h e t a )}{3} = \frac{f}{3} + \frac{2}{3}

\frac{3 cos ( ptt h e t a )}{3}

بسّط:

cos (ptt h e t a)

\frac{3 cos ( ptt h e t a )}{3}

\frac{3}{3} = 1 :

اقسم الأعداد

= cos (ptt h e t a)

\frac{f}{3} + \frac{2}{3}

بسّط:

\frac{f + 2}{3}

\frac{f}{3} + \frac{2}{3}

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

فعّل القانون

= \frac{f + 2}{3}

cos (ptt h e t a) = \frac{f + 2}{3}

cos (ptt h e t a) = \frac{f + 2}{3}

cos (ptt h e t a) = \frac{f + 2}{3}

Apply trig inverse properties

cos (ptt h e t a) = \frac{f + 2}{3}

cos (ptt h e t a) = \frac{f + 2}{3} :

حلول عامّة لـ

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

ptt h e t a = arccos (\frac{f + 2}{3}) + 2 πn, ptt h e t a = - arccos (\frac{f + 2}{3}) + 2 πn

ptt h e t a = arccos (\frac{f + 2}{3}) + 2 πn, ptt h e t a = - arccos (\frac{f + 2}{3}) + 2 πn

ptt h e t a = arccos (\frac{f + 2}{3}) + 2 πn

حلّ:

t = 3 \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

ptt h e t a = arccos (\frac{f + 2}{3}) + 2 πn

بسّط

e t^{3} p ha = arccos (\frac{f + 2}{3}) + 2 πn

e p ha; e p ha \neq = 0

اقسم الطرفين على

\frac{e t ^{3} p ha}{e p ha} = \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

t^{3} = \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

x = n f (a)

فرديّ، الحلّ هو

n, x^{n} = f (a)

لـ

t = 3 \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

ptt h e t a = - arccos (\frac{f + 2}{3}) + 2 πn

حلّ:

t = 3 \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

ptt h e t a = - arccos (\frac{f + 2}{3}) + 2 πn

بسّط

e t^{3} p ha = - arccos (\frac{f + 2}{3}) + 2 πn

e p ha; e p ha \neq = 0

اقسم الطرفين على

\frac{e t ^{3} p ha}{e p ha} = \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

t^{3} = \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

x = n f (a)

فرديّ، الحلّ هو

n, x^{n} = f (a)

لـ

t = 3 \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}; e p ha \neq = 0

t = 3 \frac{arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}, t = 3 \frac{- arccos ( \frac{f + 2}{3} ) + 2 πn}{e p ha}

رسم

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