What is the general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) ?

The general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) is x=pin-1/2 ,x= pi/2-1/2+pin

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sin(0.5pi)cos(0.5pi)=0.5sin(2x+1)

Solution

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

Solution

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

Degrees

x = - 28.64788 \dots^{\circ} + 18 0^{\circ} n, x = 61.35211 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

sin (0.5 π) = 1

sin (0.5 π)

= sin (\frac{1}{2} π)

Simplify:

\frac{1}{2} π = \frac{π}{2}

\frac{1}{2} π

Multiply fractions:

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

= \frac{1 π}{2}

Multiply:

1 π = π

= \frac{π}{2}

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

= 1

cos (0.5 π) = 0

cos (0.5 π)

= cos (\frac{1}{2} π)

Simplify:

\frac{1}{2} π = \frac{π}{2}

\frac{1}{2} π

Multiply fractions:

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

= \frac{1 π}{2}

Multiply:

1 π = π

= \frac{π}{2}

= cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0

1 \cdot 0 = 0.5 sin (2 x + 1)

Switch sides

0.5 sin (2 x + 1) = 1 \cdot 0

Apply rule

0 \cdot a = 0

0.5 sin (2 x + 1) = 0

Multiply both sides by

10

0.5 sin (2 x + 1) = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.5 sin (2 x + 1) \cdot 10 = 0 \cdot 10

Refine

5 sin (2 x + 1) = 0

5 sin (2 x + 1) = 0

Divide both sides by

5

5 sin (2 x + 1) = 0

Divide both sides by

5

\frac{5 sin ( 2 x + 1 )}{5} = \frac{0}{5}

Simplify

sin (2 x + 1) = 0

sin (2 x + 1) = 0

General solutions for

sin (2 x + 1) = 0

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}

2 x + 1 = 0 + 2 πn, 2 x + 1 = π + 2 πn

2 x + 1 = 0 + 2 πn, 2 x + 1 = π + 2 πn

Solve

2 x + 1 = 0 + 2 πn : x = πn - \frac{1}{2}

2 x + 1 = 0 + 2 πn

0 + 2 πn = 2 πn

2 x + 1 = 2 πn

Move

1

to the right side

2 x + 1 = 2 πn

Subtract

1

from both sides

2 x + 1 - 1 = 2 πn - 1

Simplify

2 x = 2 πn - 1

2 x = 2 πn - 1

Divide both sides by

2

2 x = 2 πn - 1

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2} - \frac{1}{2}

Simplify

x = πn - \frac{1}{2}

x = πn - \frac{1}{2}

Solve

2 x + 1 = π + 2 πn : x = \frac{π}{2} - \frac{1}{2} + πn

2 x + 1 = π + 2 πn

Move

1

to the right side

2 x + 1 = π + 2 πn

Subtract

1

from both sides

2 x + 1 - 1 = π + 2 πn - 1

Simplify

2 x = π + 2 πn - 1

2 x = π + 2 πn - 1

Divide both sides by

2

2 x = π + 2 πn - 1

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

Simplify

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2} : \frac{π}{2} - \frac{1}{2} + πn

\frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

Group like terms

= \frac{π}{2} - \frac{1}{2} + \frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) ?
The general solution for sin(0.5pi)cos(0.5pi)=0.5sin(2x+1) is x=pin-1/2 ,x= pi/2-1/2+pin