Question

Solve the equation for 0 ≤ x < 2π

`-cos^2(x)+sin(x)=1`

Answer 1

`-\cos^2(x)+\sin(x)=1\implies -[1-\sin^2(x)]+\sin(x)=1 \\\\\\ \sin^2(x)+\sin(x)-2=0\implies [\sin(x)]^2+\sin(x)-2=0 \\\\\\ ( ~~ \sin(x)-1 ~~ )( ~~ \sin(x)+2 ~~ )=0 \\\\[-0.35em] ~\dotfill\\\\ \sin(x)-1=0\implies \sin(x)=1\implies x=\sin^(-1)(1)\implies x=(\pi )/(2)`

now, what's wrong with the 2nd factor? sin(x) + 2 = 0?

well, we can go ahead and make it sin(x) = -2, however, let's recall that sine is never less than -1 or even more than 1, so that's out of range for sine.