Question -
Answer -
Given (sin-1 x)2 +(cos-1 x)2 = 17 π2/36
We know that cos-1 x+ sin-1 x = π/2
Then cos-1 x= π/2 – sin-1 x
Substituting this in (sin-1 x)2 +(cos-1 x)2 = 17 π2/36 we get
(sin-1 x)2 +(π/2 – sin-1 x)2 = 17 π2/36
Let y = sin-1 x
y2 +((π/2) – y)2 = 17 π2/36
y2 + π2/4– y2 – 2y ((π/2) – y) = 17 π2/36
π2/4 – πy +2 y2 = 17 π2/36
On rearranging andsimplifying, we get
2y2 –πy + 2/9 π2 = 0
18y2 –9 πy + 2 π2 = 0
18y2 –12 πy + 3 πy + 2 π2 = 0
6y (3y – 2π) + π (3y –2π) = 0
Now, (3y – 2π) = 0 and(6y + π) = 0
Therefore y = 2π/3 andy = – π/6
Now substituting y = –π/6 in y = sin-1 x we get
sin-1 x= – π/6
x = sin (- π/6)
x = -1/2
Now substituting y =-2π/3 in y = sin-1 x we get
x = sin (2π/3)
x = √3/2
Now substituting x =√3/2 in (sin-1 x)2 + (cos-1 x)2 =17 π2/36 we get,
= π/3 + π/6
= π/2 which is notequal to 17 π2/36
So we have to neglectthis root.
Now substituting x =-1/2 in (sin-1 x)2 + (cos-1 x)2 =17 π2/36 we get,
= π2/36 + 4π2/9
= 17 π2/36
Hence x = -1/2.