Question

solve x minus a upon x minus b + x minus b upon x minus A is equals to A upon B + b Upon A

Answer 1

Solution:

We have,

2 sin x - 1 = 0

⇒ sin x = ½

⇒ sin x = sin π6π6

⇒ sin x = sin (π  - π6π6)

⇒ sin x =  sin 5π65π6 



 

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, π2π2, π, 3π23π2, 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 π6π6

⇒ sin x = sin (2nπ + π6π6), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Therefore, x = 2nπ + π6π6 …………….. (i)

Again, if the final arm OP of the unit circle lies in the second quadrant, then        

sin x = ½

⇒ sin x = sin 5π65π6 

⇒ sin x = sin (2nπ + 5π65π6), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Therefore, x = 2nπ + 5π65π6 …………….. (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)22 π6π6, n ∈ I

Answer 2

Answer:

Step-by-step explanation:

Please see attachment for answer.

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