Question

2(sin^-1x)² - 5sin^-1x + 2 =0​

Answer 1

Answer:

put sin^-1x =t

the equation then becomes 2t^2-5t+2=0

now by factorizing we get:

2t^2-4t-t+2=0

2t(t-2)+(-1)(t-2)=0

(t-2)(2t-1)=0

now cases:

case 1:

t=2 implies sin^-1(x)=2

but this is impossible since range of sin^-1x is

[-π/2, π/2]

case 2:

t=1/2 implies sin^-1x=1/2

x=sin(1/2)

is the solution