Question -
Answer -
Given (sin-1┬аx)2┬а+(cos-1┬аx)2┬а= 17 ╧А2/36
We know that cos-1┬аx+ sin-1┬аx = ╧А/2
Then cos-1┬аx= ╧А/2 тАУ sin-1┬аx
Substituting this in (sin-1┬аx)2┬а+(cos-1┬аx)2┬а= 17 ╧А2/36 we get
(sin-1┬аx)2┬а+(╧А/2 тАУ sin-1┬аx)2┬а= 17 ╧А2/36
Let y = sin-1┬аx
y2┬а+((╧А/2) тАУ y)2┬а= 17 ╧А2/36
y2┬а+ ╧А2/4тАУ y2┬атАУ 2y ((╧А/2) тАУ y) = 17 ╧А2/36
╧А2/4 тАУ ╧Аy +2 y2┬а= 17 ╧А2/36
On rearranging andsimplifying, we get
2y2┬атАУ╧Аy + 2/9 ╧А2┬а= 0
18y2┬атАУ9 ╧Аy + 2 ╧А2┬а= 0
18y2┬атАУ12 ╧Аy + 3 ╧Аy + 2 ╧А2┬а= 0
6y (3y тАУ 2╧А) + ╧А (3y тАУ2╧А) = 0
Now, (3y тАУ 2╧А) = 0 and(6y + ╧А) = 0
Therefore y = 2╧А/3 andy = тАУ ╧А/6
Now substituting y = тАУ╧А/6 in y = sin-1┬аx we get
sin-1┬аx= тАУ ╧А/6
x = sin (- ╧А/6)
x = -1/2
Now substituting y =-2╧А/3 in y = sin-1┬аx we get
x = sin (2╧А/3)
x = тИЪ3/2
Now substituting x =тИЪ3/2 in (sin-1┬аx)2┬а+ (cos-1┬аx)2┬а=17 ╧А2/36 we get,
= ╧А/3 + ╧А/6
= ╧А/2 which is notequal to 17 ╧А2/36
So we have to neglectthis root.
Now substituting x =-1/2 in (sin-1┬аx)2┬а+ (cos-1┬аx)2┬а=17 ╧А2/36 we get,
= ╧А2/36 + 4╧А2/9
= 17 ╧А2/36
Hence x = -1/2.