# 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
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.

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:
$$\displaystyle{x}=\frac{\pi}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{5}\pi}}{{4}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{3}\pi}}{{2}}+{2}{k}\pi{\quad\text{or}\quad}{x}=\frac{{{7}\pi}}{{4}}+{2}{k}\pi$$

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