Question

Find the exact solutions of the equation that are in the interval [0, 2π ).

sin 2t + sin t = 0

Answer 1

\[\large \sin(2t) + \sin(t) = 0 \implies \sin(2t) = -\sin(t) \implies \sin(2t) = \sin(-t)\]

So:

general solution is given by:

\[\huge 2t = n \pi + (-1)^n \cdot (-t)\]

Put n = 0 here:

\[\large 2t = -t \implies t = 0\]

Answer 2

Like wise put n= 1,

\[\large 2t = \pi + (-)(-t) \implies 2t = \pi + t \implies t = \pi\]

Answer 3

Similarly put n= 2:

\[\large 2t = 2 \pi - t \implies t = \frac{2 \pi}{3}\]