Theorem 1

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General solution of sin q = k

Method I

Given sin q = k .......(i)

Let a be any angle, such that sin a = k.

(In actual practice, we choose the smallest positive angle such that sin a = k.)

From (i) and (ii), we have

Method II

Given equation sin q = k.

Let a be the least positive angle, such that sin a = k.

Also, any angle co-terminal with the angles a, p - a is trigonometrically equivalent and so has the same 'sine'.

Particular Cases

i) If sin q = 0, then a = 0.

Hence, we get q = np + (-1)ⁿ 0 = np.