Question

In degrees, what are all values of 'x' between 0 and 360 for which \[\sin x > \sqrt{1-\sin^2 x}\]?

Answer 1

right hand side is
\[|\cos(x)|\] if that helps

Answer 2

i believe your answer (in degrees) is 45 degrees to 135 decrees

Answer 3

don't forget that
\[\sqrt{1-\sin^2(x)}=|\cos(x)|\] and not
\[\cos(x)\]

Answer 4

how does that lead you to the solution?

Answer 5

satellite is correct in using the absolute value....

Answer 6

how do you square both sides in inequality?