solve for x cos(2sin-1x) =1/9 ,x 0 - Maths - Inverse Trigonometric Functions

Question

solve for x
cos(2sin- 1x) =1/9 ,x>0

Gursheen kaur. · Accepted Answer

Let sin- 1x=y â‡’x = siny (i) Now, cosy = $\sqrt{1 - \sin^{2} x} = \sqrt{1 - x^{2}} (i i)$ Thus, the given expression will become cos(2sin- 1x)=1/9 â‡’cos2y=1/9 $\Rightarrow \cos^{2} y - \sin^{2} y = \frac{1}{9} [∵ \cos 2 x = \cos^{2} x - \sin^{2} x] \Rightarrow 1 - x^{2} - x^{2} = \frac{1}{9} [F r o m (i) a n d (i i)] \Rightarrow 2 x^{2} = 1 - \frac{1}{9} \Rightarrow 2 x^{2} = \frac{8}{9} \Rightarrow x^{2} = \frac{4}{9} \Rightarrow x = \frac{2}{3}$

solve for xcos(2sin- 1x) =1/9 ,x>0

solve for x
cos(2sin^- ¹x) =1/9 ,x>0