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}

Solution steps

sin (2 p + 1) = - 1

General solutions for

sin (2 p + 1) = - 1

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}

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