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

Question
Equations and inequalities
asked 2021-02-12
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.
\(\displaystyle{\cos{{2}}}{u}+{\sin{{u}}}={0}\)

Answers (1)

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 \(\displaystyle{321.82}^{{\circ}}\)
0

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