What is the general solution for sin(2p+1)=-1 ?

The general solution for sin(2p+1)=-1 is p=pin-1/2+(3pi)/4

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sin(2p+1)=-1

sin (2 p + 1) = - 1

p = πn - \frac{1}{2} + \frac{3 π}{4}

Degrees

p = 106.35211 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 p + 1) = - 1

General solutions for

sin (2 p + 1) = - 1

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 p + 1 = \frac{3 π}{2} + 2 πn

2 p + 1 = \frac{3 π}{2} + 2 πn

Solve

2 p + 1 = \frac{3 π}{2} + 2 πn : p = πn - \frac{1}{2} + \frac{3 π}{4}

2 p + 1 = \frac{3 π}{2} + 2 πn

Move

1

to the right side

2 p + 1 = \frac{3 π}{2} + 2 πn

Subtract

1

from both sides

2 p + 1 - 1 = \frac{3 π}{2} + 2 πn - 1

Simplify

2 p = \frac{3 π}{2} + 2 πn - 1

2 p = \frac{3 π}{2} + 2 πn - 1

Divide both sides by

2

2 p = \frac{3 π}{2} + 2 πn - 1

Divide both sides by

2

\frac{2 p}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} - \frac{1}{2}

Simplify

\frac{2 p}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} - \frac{1}{2}

Simplify

\frac{2 p}{2} : p

\frac{2 p}{2}

Divide the numbers:

\frac{2}{2} = 1

= p

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} - \frac{1}{2} : πn - \frac{1}{2} + \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} - \frac{1}{2}

Group like terms

= - \frac{1}{2} + \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

= - \frac{1}{2} + πn + \frac{3 π}{4}

Group like terms

= πn - \frac{1}{2} + \frac{3 π}{4}

p = πn - \frac{1}{2} + \frac{3 π}{4}

p = πn - \frac{1}{2} + \frac{3 π}{4}

p = πn - \frac{1}{2} + \frac{3 π}{4}

p = πn - \frac{1}{2} + \frac{3 π}{4}

y, a \cdot z = 5 sin (2 y)

sin^{3} (x) = sin^{2} (x)

2 sec^{2} (a) + tan^{2} (a) = 3

3 cos^{2} (x) + 4 cos (x) + 1 = 0

cos^{5} (x) = \frac{1}{2}

What is the general solution for sin(2p+1)=-1 ?
The general solution for sin(2p+1)=-1 is p=pin-1/2+(3pi)/4