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

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

anonymous · Answer

please answer at least one! (:

mr.math · Answer

First solve them as quadratic equations.

mr.math · Answer

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.