sin^2x-2sinx-1=0
This is just a quadratic in sinx:
Let u=sinx
Then:
u^2-2u-1=0
Using the quadratic formula:
(-(-2)+-sqrt((-2)^2-4(1)(-1)))/(2(1)
(2+-sqrt(4+4))/(2)
(2+-sqrt(8))/(2)
(2+-2sqrt(2))/(2) = 1+sqrt(2) , color(white)(888)1-sqrt(2)
u=sinx
:.
sinx=1+sqrt(2), color(white)(888)sinx=1-sqrt(2)
x=arcsin(sinx)=arcsin(1+sqrt(2)) ( no real solutions )*
x=arcsin(sinx)=arcsin(1-sqrt(2))=arcsin(1-sqrt(2))
2pi + arcsin(1-sqrt(2))color(white)(88) IV quadrant
pi-arcsin(1-sqrt(2))color(white)(888) III quadrant
These values are in the interval:
0<=x<=2pi
For the general solution:
pi-arcsin(1-sqrt(2))+npi
4pin+arcsin(1-sqrt(2))
For integers bbn
*
domain of arcsin(x)
[-1,1]
1+sqrt(2)~~2.414213562
