Find the solution to this equation: displaystylesqrt{{2}} cos{{left({x}right)}} sin{{left({x}right)}}+ cos{{left({x}right)}}={0} The solution should be such that all angles are in radian. for solution the first angle should be between displaystyle{left[{0},{2}piright)} and then the period. And when 2 or more solutions are available then the solution must be in increasing order of the angles.

Find the solution to this equation: displaystylesqrt{{2}} cos{{left({x}right)}} sin{{left({x}right)}}+ cos{{left({x}right)}}={0} The solution should be such that all angles are in radian. for solution the first angle should be between displaystyle{left[{0},{2}piright)} and then the period. And when 2 or more solutions are available then the solution must be in increasing order of the angles.

Question
Decimals
asked 2021-01-31
Find the solution to this equation:
\(\displaystyle\sqrt{{2}} \cos{{\left({x}\right)}} \sin{{\left({x}\right)}}+ \cos{{\left({x}\right)}}={0}\)
The solution should be such that all angles are in radian. for solution the first angle should be between \(\displaystyle{\left[{0},{2}\pi\right)}\) and then the period.
And when 2 or more solutions are available then the solution must be in increasing order of the angles.

Answers (1)

2021-02-01
Step 1
\(\displaystyle\sqrt{{2}} \cos{{\left({x}\right)}} \sin{{\left({x}\right)}}+ \cos{{\left({x}\right)}}={0}\)
\(\displaystyle \cos{{\left({x}\right)}}{\left(\sqrt{{2}} \sin{{\left({x}\right)}}+{1}\right)}={0}\)
\(\displaystyle \cos{{\left({x}\right)}}={0}{\quad\text{or}\quad}\sqrt{{2}} \sin{{\left({x}\right)}}+{1}={0}\)
\(\displaystyle \cos{{\left({x}\right)}}={0}{\quad\text{or}\quad} \sin{{\left({x}\right)}}=-\frac{1}{\sqrt{{2}}}\)
first case:
when \(\displaystyle \cos{{x}}={0}\).
\(\displaystyle \cos{{x}}= \cos{{\left(\frac{\pi}{{2}}\right)}}\)
therefore, \(\displaystyle{x}=\frac{\pi}{{2}}+{2}{k}\pi\)
Step 2
when \(\displaystyle \cos{{x}}={0}.\)
\(\displaystyle \cos{{x}}= \cos{{\left(\frac{{{3}\pi}}{{2}}\right)}}\)
therefore, \(\displaystyle{x}=\frac{{{3}\pi}}{{2}}+{2}{k}\pi\)
second case:
when \(\displaystyle \sin{{x}}=\frac{{-{1}}}{\sqrt{{2}}}\)
\(\displaystyle \sin{{x}}= \sin{{\left(\frac{{{5}\pi}}{{4}}\right)}}\)
therefore, \(\displaystyle{x}=\frac{{{5}\pi}}{{4}}+{2}{k}\pi\)
Step 3
when \(\displaystyle \sin{{x}}=\frac{{-{1}}}{\sqrt{{2}}}\)
\(\displaystyle \sin{{x}}= \sin{{\left(\frac{{{7}\pi}}{{4}}\right)}}\)
therefore, \(\displaystyle{x}=\frac{{{7}\pi}}{{4}}+{2}{k}\pi\)
therefore the combined solution is:
\(\displaystyle{x}=\frac{\pi}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{5}\pi}}{{4}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{3}\pi}}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{7}\pi}}{{4}}+{2}{k}\pi\)
Step 4
therefore the solution of the given equation \(\displaystyle\sqrt{{2}} \cos{{\left({x}\right)}} \sin{{\left({x}\right)}}+ \cos{{\left({x}\right)}}={0}\) is:
\(\displaystyle{x}=\frac{\pi}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{5}\pi}}{{4}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{3}\pi}}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{7}\pi}}{{4}}+{2}{k}\pi\)
0

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