What is the general solution for sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2 ?

Q: What is the general solution for sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2 ?

The general solution for sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2 is t=(23pi)/(12),t=(31pi)/(12)

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sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2

Solution

sin (t + \frac{π}{4}) = \frac{1}{2}, π \leq t \leq \frac{7 π}{2}

Solution

t = \frac{23 π}{12}, t = \frac{31 π}{12}

Degrees

t = 34 5^{\circ}, t = 46 5^{\circ}

Solution steps

sin (t + \frac{π}{4}) = \frac{1}{2}, π \leq t \leq \frac{7 π}{2}

General solutions for

sin (t + \frac{π}{4}) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t + \frac{π}{4} = \frac{π}{6} + 2 πn, t + \frac{π}{4} = \frac{5 π}{6} + 2 πn

t + \frac{π}{4} = \frac{π}{6} + 2 πn, t + \frac{π}{4} = \frac{5 π}{6} + 2 πn

Solve

t + \frac{π}{4} = \frac{π}{6} + 2 πn : t = 2 πn - \frac{π}{12}

t + \frac{π}{4} = \frac{π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

t + \frac{π}{4} = \frac{π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

t + \frac{π}{4} - \frac{π}{4} = \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

t + \frac{π}{4} - \frac{π}{4} = \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

t + \frac{π}{4} - \frac{π}{4} : t

t + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= t

Simplify

\frac{π}{6} + 2 πn - \frac{π}{4} : 2 πn - \frac{π}{12}

\frac{π}{6} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{6} - \frac{π}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 2}{12} - \frac{π 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π 3}{12}

Add similar elements:

2 π - 3 π = - π

= \frac{- π}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= 2 πn - \frac{π}{12}

t = 2 πn - \frac{π}{12}

t = 2 πn - \frac{π}{12}

t = 2 πn - \frac{π}{12}

Solve

t + \frac{π}{4} = \frac{5 π}{6} + 2 πn : t = 2 πn + \frac{7 π}{12}

t + \frac{π}{4} = \frac{5 π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

t + \frac{π}{4} = \frac{5 π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

t + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

t + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{6} + 2 πn - \frac{π}{4}

Simplify

t + \frac{π}{4} - \frac{π}{4} : t

t + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= t

Simplify

\frac{5 π}{6} + 2 πn - \frac{π}{4} : 2 πn + \frac{7 π}{12}

\frac{5 π}{6} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{5 π}{6}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

4

6

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

For

\frac{5 π}{6} :

multiply the denominator and numerator by

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= - \frac{π 3}{12} + \frac{10 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 3 + 10 π}{12}

Add similar elements:

- 3 π + 10 π = 7 π

= 2 πn + \frac{7 π}{12}

t = 2 πn + \frac{7 π}{12}

t = 2 πn + \frac{7 π}{12}

t = 2 πn + \frac{7 π}{12}

t = 2 πn - \frac{π}{12}, t = 2 πn + \frac{7 π}{12}

Solutions for the range

π \leq t \leq \frac{7 π}{2}

t = \frac{23 π}{12}, t = \frac{31 π}{12}

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Frequently Asked Questions (FAQ)

What is the general solution for sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2 ?
The general solution for sin(t+pi/4)= 1/2 ,pi<= t<= (7pi)/2 is t=(23pi)/(12),t=(31pi)/(12)