# 2 sin x - 1 = 0

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

Trigonometric Equations

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