Question Number 78885 by gopikrishnan last updated on 21/Jan/20
![The number of real numbers in [0,2pi] satisfying sin^(−1) x−2sin^2 x+1=0 is](https://www.tinkutara.com/question/Q78885.png)
$${The}\:{number}\:{of}\:{real}\:{numbers}\:{in}\:\left[\mathrm{0},\mathrm{2pi}\right]\:\mathrm{satisfying}\:\mathrm{sin}^{−\mathrm{1}} \mathrm{x}−\mathrm{2sin}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{0}\:{is} \\ $$
Answered by mind is power last updated on 21/Jan/20
![sin^− (x) is defind in[−1,1] sir](https://www.tinkutara.com/question/Q78893.png)
$${sin}^{−} \left({x}\right)\:{is}\:{defind}\:{in}\left[−\mathrm{1},\mathrm{1}\right]\:{sir} \\ $$