Question

# Solving the following equations will require you to use the quadratic formula. Solve each equation for u between 0^(circ) and 360^(circ), and round your answers to the nearest tenth of a degree. cos2 u + sin u = 0

Equations and inequalities
Solving the following equations will require you to use the quadratic formula. Solve each equation for u between $$\displaystyle{0}^{{\circ}}{\quad\text{and}\quad}{360}^{{\circ}}$$, and round your answers to the nearest tenth of a degree.
$$\displaystyle{\cos{{2}}}{u}+{\sin{{u}}}={0}$$

2021-02-13
We have,
$$\displaystyle{{\cos}^{{{2}}}{u}}+{\sin{{u}}}={0}$$
By using the identity, $$\displaystyle{{\cos}^{{{2}}}{u}}={1}−{\sin{{2}}}{u}$$ and solving further,
$$\displaystyle{{\cos}^{{{2}}}{u}}+{\sin{{u}}}={0}$$
$$\displaystyle{1}−{{\sin}^{{{2}}}{u}}+{\sin{=}}{0}$$
$$\displaystyle{{\sin}^{{{2}}}{u}}−{\sin{{u}}}−{1}={0}$$
The above equation is quadratic equation in variable sinu.
Using the quadratic equation formula and solving it further, we get our result as
$$\displaystyle{\sin{{u}}}={\frac{{-{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({1}\right)}}}}$$
$$\displaystyle{\sin{{u}}}={\frac{{{1}+\sqrt{{{5}}}}}{{{2}}}},{\frac{{{1}-\sqrt{{{5}}}}}{{{2}}}}$$
Now,
$$\displaystyle{\sin{{u}}}={\frac{{{1}+\sqrt{{{5}}}}}{{{2}}}}$$
No solution occurs because range for sine function lies between -1 to 1
For,
$$\displaystyle{\sin{{u}}}={\frac{{{1}-\sqrt{{{5}}}}}{{{2}}}}$$
$$\displaystyle{\sin{{u}}}={0.618}$$
$$\displaystyle{u}={{\sin}^{{-{1}}}{\left(-{0.618}\right)}}$$
$$\displaystyle{u}=-{{\sin}^{{-{1}}}{\left({0.618}\right)}}$$
$$\displaystyle{u}={38.17}^{{\circ}}$$
$$\displaystyle{u}={360}^{{\circ}}-{38.17}^{{\circ}}$$
$$\displaystyle{u}={321.82}^{{\circ}}$$
Hence,value of u will be $$\displaystyle{321.82}^{{\circ}}$$