Question

solve for x over the interval [0,2π)

1) sin²x=4sinx+6

2) 2cos²x+sinx-1 =0

3) 4sec² (x/2) - 5sec(x/2) -6 =0

Answer 1

please answer at least one! (:

Answer 2

First solve them as quadratic equations.

Answer 3

I will help you with the first one. You can notice that \(\sin²x-4\sin{x}-6=0\) is quadratic in \(\sin{x}\), we can apply the quadratic formula and find
\[\sin{x}=\frac{4\pm \sqrt{16+24}}{2}=2\pm \sqrt{10}.\]
But both \(2\pm \sqrt{10}\) are not in the range of \(\sin{x}\), hence the equation has NO real solutions.