Question

Find the principal solution of the following equation:

sin θ = `-1/2`

Answer 1

We now that,
`sin  (pi)/(6) = (1)/(2) and sin(pi + θ)` = – sin θ,
sin(2π - θ) = – sinθ.

∴ `sin(pi + pi/6) = -sin pi/(6) = -(1)/(2)`

`"and" sin(2pi - pi/6) = - sin  pi/6 = - (1)/(2)`

∴ `sin  (7pi)/(6) = sin  (11pi)/(6) = -(1)/(2)`, where

`0 < (7pi)/(6) < 2pi and 0 < (11pi)/(6) < 2pi`

∴ sinθ = `-(1)/(2)"gives"`,

sinθ = `sin  (7pi)/(6) = sin  (11pi)/(6)`

∴ θ = `(7pi)/(6) and θ = (11pi)/(6)`
Hence, the required principal solutions are

θ = `(7pi)/(6) and θ = (11pi)/(6)`.