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Popular >

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

Solution

solve for

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

Solution

Solution steps

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

Switch sides

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

Move

2

to the right side

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

Add

2

to both sides

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

Simplify

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

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

Divide both sides by

3

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

Divide both sides by

3

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

Simplify

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

Simplify

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

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

Divide the numbers:

\frac{3}{3} = 1

= cos (ptt h e t a)

Simplify

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

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

Apply rule

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

= \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}

General solutions for

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

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

Solve

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

Simplify

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

Divide both sides by

e p ha; e p ha \neq = 0

\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

For

x^{n} = f (a)

, n is odd, the solution is

Solve

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

Simplify

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

Divide both sides by

e p ha; e p ha \neq = 0

\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

For

x^{n} = f (a)

, n is odd, the solution is

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