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प्रश्न
Find the principal solution of the following equation:
sin θ = `-1/2`
उत्तर
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)`.
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