What is the general solution for solvefor h=7cos(pi/3 t),t ?

The general solution for solvefor h=7cos(pi/3 t),t is t=(3arccos(h/7))/pi+6n,t=-(3arccos(h/7))/pi+6n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor h=7cos(pi/3 t),t

Solution

solve for

h = 7 cos (\frac{π}{3} t), t

Solution

t = \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n, t = - \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

Solution steps

h = 7 cos (\frac{π}{3} t)

Switch sides

7 cos (\frac{π}{3} t) = h

Divide both sides by

7

7 cos (\frac{π}{3} t) = h

Divide both sides by

7

\frac{7 cos ( \frac{π}{3} t )}{7} = \frac{h}{7}

Simplify

cos (\frac{π}{3} t) = \frac{h}{7}

cos (\frac{π}{3} t) = \frac{h}{7}

Apply trig inverse properties

cos (\frac{π}{3} t) = \frac{h}{7}

General solutions for

cos (\frac{π}{3} t) = \frac{h}{7}

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

\frac{π}{3} t = arccos (\frac{h}{7}) + 2 πn, \frac{π}{3} t = - arccos (\frac{h}{7}) + 2 πn

\frac{π}{3} t = arccos (\frac{h}{7}) + 2 πn, \frac{π}{3} t = - arccos (\frac{h}{7}) + 2 πn

Solve

\frac{π}{3} t = arccos (\frac{h}{7}) + 2 πn : t = \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

\frac{π}{3} t = arccos (\frac{h}{7}) + 2 πn

Multiply both sides by

3

\frac{π}{3} t = arccos (\frac{h}{7}) + 2 πn

Multiply both sides by

3

3 \cdot \frac{π}{3} t = 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Simplify

3 \cdot \frac{π}{3} t = 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Simplify

3 \cdot \frac{π}{3} t : π t

3 \cdot \frac{π}{3} t

Multiply fractions:

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

= \frac{3 π}{3} t

Cancel the common factor:

3

= t π

Simplify

3 arccos (\frac{h}{7}) + 3 \cdot 2 πn : 3 arccos (\frac{h}{7}) + 6 πn

3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= 3 arccos (\frac{h}{7}) + 6 πn

π t = 3 arccos (\frac{h}{7}) + 6 πn

π t = 3 arccos (\frac{h}{7}) + 6 πn

π t = 3 arccos (\frac{h}{7}) + 6 πn

Divide both sides by

π

π t = 3 arccos (\frac{h}{7}) + 6 πn

Divide both sides by

π

\frac{π t}{π} = \frac{3 arccos ( \frac{h}{7} )}{π} + \frac{6 πn}{π}

Simplify

t = \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

t = \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

Solve

\frac{π}{3} t = - arccos (\frac{h}{7}) + 2 πn : t = - \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

\frac{π}{3} t = - arccos (\frac{h}{7}) + 2 πn

Multiply both sides by

3

\frac{π}{3} t = - arccos (\frac{h}{7}) + 2 πn

Multiply both sides by

3

3 \cdot \frac{π}{3} t = - 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Simplify

3 \cdot \frac{π}{3} t = - 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Simplify

3 \cdot \frac{π}{3} t : π t

3 \cdot \frac{π}{3} t

Multiply fractions:

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

= \frac{3 π}{3} t

Cancel the common factor:

3

= t π

Simplify

- 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn : - 3 arccos (\frac{h}{7}) + 6 πn

- 3 arccos (\frac{h}{7}) + 3 \cdot 2 πn

Multiply the numbers:

3 \cdot 2 = 6

= - 3 arccos (\frac{h}{7}) + 6 πn

π t = - 3 arccos (\frac{h}{7}) + 6 πn

π t = - 3 arccos (\frac{h}{7}) + 6 πn

π t = - 3 arccos (\frac{h}{7}) + 6 πn

Divide both sides by

π

π t = - 3 arccos (\frac{h}{7}) + 6 πn

Divide both sides by

π

\frac{π t}{π} = - \frac{3 arccos ( \frac{h}{7} )}{π} + \frac{6 πn}{π}

Simplify

t = - \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

t = - \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

t = \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n, t = - \frac{3 arccos ( \frac{h}{7} )}{π} + 6 n

Popular Examples

3+3sin(θ)=3+3cos(θ)

3 + 3 sin (θ) = 3 + 3 cos (θ)

-sqrt(3)=tan(θ)

- 3 = tan (θ)

cot(θ)=13

cot (θ) = 13

arctan(0.2x)+arctan(0.0625x)=54

arctan (0.2 x) + arctan (0.0625 x) = 5 4^{\circ}

sin^2(θ)+cos(θ)=1

sin^{2} (θ) + cos (θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for solvefor h=7cos(pi/3 t),t ?
The general solution for solvefor h=7cos(pi/3 t),t is t=(3arccos(h/7))/pi+6n,t=-(3arccos(h/7))/pi+6n