Solution 4.4:6b

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After moving the terms over to the left-hand side, so that
After moving the terms over to the left-hand side, so that
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{{Displayed math||<math>\sqrt{2}\sin x\cos x-\cos x=0</math>}}
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<math>\sqrt{2}\sin x\cos x-\cos x=0</math>
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we see that we can take out a common factor <math>\cos x</math>,
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{{Displayed math||<math>\cos x (\sqrt{2}\sin x-1) = 0</math>}}
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we see that we can take out a common factor
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and that the equation is only satisfied if at least one of the factors <math>\cos x</math> or <math>\sqrt{2}\sin x - 1</math> is zero. Thus, there are two cases:
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<math>\text{cos }x</math>,
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<math>\cos x\left( \sqrt{2}\sin x-1 \right)=0</math>
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<math>\cos x=0:</math>
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This basic equation has solutions <math>x=\pi/2</math> and <math>x=3\pi/2</math> in the unit circle, and from this we see that the general solution is
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and that the equation is only satisfied if at least one of the factors,
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{{Displayed math||<math>x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,</math>}}
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<math>\text{cos }x</math>
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or
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<math>\sqrt{2}\text{sin }x-\text{1}</math>
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is zero. Thus, there are two cases:
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 +
where ''n'' is an arbitrary integer. Because the angles <math>\pi/2</math> and
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<math>3\pi/2</math> differ by <math>\pi</math>, the solutions can be summarized as
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<math>\text{cos }x=0</math>: This basic equation has solutions
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{{Displayed math||<math>x=\frac{\pi}{2}+n\pi\,,</math>}}
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<math>x={\pi }/{2}\;</math>
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and
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<math>x=3{\pi }/{2}\;</math>
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in the unit circle, and from this we see that the general solution is
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where ''n'' is an arbitrary integer.
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<math>x=\frac{\pi }{2}+2n\pi </math>
 
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and
 
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<math>x=\frac{3\pi }{2}+2n\pi </math>
 
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<math>\sqrt{2}\sin x - 1 = 0:</math>
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where
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If we rearrange the equation, we obtain the basic equation as <math>\sin x = 1/\!\sqrt{2}</math>, which has the solutions <math>x=\pi/4</math> and <math>x=3\pi /4</math> in the unit circle and hence the general solution
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<math>n\text{ }</math>
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is an arbitrary integer. Because the angles
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<math>{\pi }/{2}\;</math>
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and
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<math>3{\pi }/{2}\;</math>
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differ by
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<math>\pi </math>, the solutions can be summarized as
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{{Displayed math||<math>x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,</math>}}
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<math>x=\frac{\pi }{2}+n\pi </math>
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(
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<math>n</math>
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an arbitrary integer).
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√
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where ''n'' is an arbitrary integer.
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<math>\sqrt{2}\text{sin }x-\text{1}=0</math>
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: if we rearrange the equation, we obtain the basic equation as
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<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the solutions
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<math>x={\pi }/{4}\;</math>
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and
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<math>x=3{\pi }/{4}\;</math>
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in the unit circle and hence the general solution
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- 
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<math>x=\frac{\pi }{4}+2n\pi </math>
 
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and
 
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<math>x=\frac{3\pi }{4}+2n\pi </math>
 
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- 
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where
 
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<math>n\text{ }</math>
 
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can arbitrary integer.
 
All in all, the original equation has the solutions
All in all, the original equation has the solutions
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{{Displayed math||<math>\left\{\begin{align}
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x &= \frac{\pi}{4}+2n\pi\,,\\[5pt]
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x &= \frac{\pi}{2}+n\pi\,,\\[5pt]
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x &= \frac{3\pi}{4}+2n\pi\,,
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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where ''n'' is an arbitrary integer.
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x=\frac{\pi }{4}+2n\pi \\
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x=\frac{\pi }{2}+n\pi \\
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x=\frac{3\pi }{4}+2n\pi \\
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\end{array} \right.</math>
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(
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<math>n\text{ }</math>
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an arbitrary integer).
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Current revision

After moving the terms over to the left-hand side, so that

\displaystyle \sqrt{2}\sin x\cos x-\cos x=0

we see that we can take out a common factor \displaystyle \cos x,

\displaystyle \cos x (\sqrt{2}\sin x-1) = 0

and that the equation is only satisfied if at least one of the factors \displaystyle \cos x or \displaystyle \sqrt{2}\sin x - 1 is zero. Thus, there are two cases:


\displaystyle \cos x=0:

This basic equation has solutions \displaystyle x=\pi/2 and \displaystyle x=3\pi/2 in the unit circle, and from this we see that the general solution is

\displaystyle x=\frac{\pi}{2}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{2}+2n\pi\,,

where n is an arbitrary integer. Because the angles \displaystyle \pi/2 and \displaystyle 3\pi/2 differ by \displaystyle \pi, the solutions can be summarized as

\displaystyle x=\frac{\pi}{2}+n\pi\,,

where n is an arbitrary integer.


\displaystyle \sqrt{2}\sin x - 1 = 0:

If we rearrange the equation, we obtain the basic equation as \displaystyle \sin x = 1/\!\sqrt{2}, which has the solutions \displaystyle x=\pi/4 and \displaystyle x=3\pi /4 in the unit circle and hence the general solution

\displaystyle x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,

where n is an arbitrary integer.


All in all, the original equation has the solutions

\displaystyle \left\{\begin{align}

x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{\pi}{2}+n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.

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