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We will discuss about the general solution of the equation 2 sin x minus 1 equals 0 (i.e., 2 sin x - 1 = 0) or sin x equals half (i.e., sin x = ½).
How to find the general solution of the trigonometric equation sin x = ½ or 2 sin x - 1 = 0?
Solution:
We have,
2 sin x - 1 = 0
⇒ sin x = ½
⇒ sin x = sin π6
⇒ sin x = sin (π - π6)
⇒ sin x = sin 5π6
Let O be the center of a unit circle. We know that in unit
circle, the length of the circumference is 2π.
If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, π2, π, 3π2, and 2π.
Therefore, from the above unit circle it is clear that the final arm OP of the angle x lies either in the first or in the second.
If the final arm OP of the unit circle lies in the first quadrant, then
sin x = ½
⇒ sin x = sin π6
⇒ sin x = sin (2nπ + π6), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = 2nπ + π6 …………….. (i)
Again, if the final arm OP of the unit circle lies in the second quadrant, then
sin x = ½
⇒ sin x = sin 5π6
⇒ sin x = sin (2nπ + 5π6), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = 2nπ + 5π6 …………….. (ii)
Therefore, the general solution of equation sin x = ½ or 2 sin x - 1 = 0 are the infinite sets of value of x given in (i) and (ii).
Hence general solution of 2 sin x - 1 = 0 is x = nπ + (-1)2 π6, n ∈ I
11 and 12 Grade Math
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