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 \(\frac{π}{6}\)
⇒ sin x = sin (π  \(\frac{π}{6}\))
⇒ sin x = sin \(\frac{5π}{6}\)
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, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), 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 \(\frac{π}{6}\)
⇒ sin x = sin (2nπ + \(\frac{π}{6}\)), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = 2nπ + \(\frac{π}{6}\) …………….. (i)
Again, if the final arm OP of the unit circle lies in the second quadrant, then
sin x = ½
⇒ sin x = sin \(\frac{5π}{6}\)
⇒ sin x = sin (2nπ + \(\frac{5π}{6}\)), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = 2nπ + \(\frac{5π}{6}\) …………….. (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)\(^{2}\) \(\frac{π}{6}\), n ∈ I
11 and 12 Grade Math
From 2 sin x Minus 1 Equals 0 to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.