Question

How do you solve `9+sin^2x=10` for `0<=x<=2pi`?

Answer 1

`color(blue)(pi/2 , (3pi)/2)`

Explanation:

`9+sin^2x=10`

Subtract 9 from both sides:

`sin^2x=1`

`sinx=+-sqrt(1)=+-1`

`x=arcsin(sinx)=arcsin(1)=>x=color(blue)(pi/2)`

`x=arcsin(sinx)=arcsin(-1)=>x=color(blue)((3pi)/2 )`