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

The general solution for sin(2pi+x)-sin(2pi-x)=-1 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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sin(2pi+x)-sin(2pi-x)=-1

sin (2 π + x) - sin (2 π - x) = - 1

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

sin (2 π + x) - sin (2 π - x) = - 1

Rewrite using trig identities

2 sin (x) = - 1

Divide both sides by

2

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

θ, ma = T sin (θ) - m g

sin (x) = \frac{16}{20}

tan (\frac{π}{4} + x) = 2 + 3

3 cos (x) = 2 sin (x)

2 cos^{2} (θ) - cos (θ) = 1, 0 \leq θ \leq 2 π

What is the general solution for sin(2pi+x)-sin(2pi-x)=-1 ?
The general solution for sin(2pi+x)-sin(2pi-x)=-1 is x=(7pi)/6+2pin,x=(11pi)/6+2pin