Solution 4.4:8a

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If we use the formula for double angles,
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If we use the formula for double angles, <math>\sin 2x = 2\sin x\cos x</math>, and move all the terms over to the left-hand side, the equation becomes
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<math>\text{sin 2}x=\text{2sin }x\text{ cos }x</math>, and move all the terms over to the left-hand side, the equation becomes
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{{Displayed math||<math>2\sin x\cos x-\sqrt{2}\cos x=0\,\textrm{.}</math>}}
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<math>2\sin x\cos x-\sqrt{2}\cos x=0.</math>
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Then, we see that we can take a factor <math>\cos x</math> out of both terms,
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{{Displayed math||<math>\cos x\,(2\sin x-\sqrt{2}) = 0</math>}}
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Then, we see that we can take a factor cos x out of both terms,
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and hence divide up the equation into two cases. The equation is satisfied either if <math>\cos x = 0</math> or if <math>2\sin x-\sqrt{2} = 0\,</math>.
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<math>\cos x\left( 2\sin x-\sqrt{2} \right)=0</math>
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<math>\cos x = 0</math>:
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This equation has the general solution
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and hence divide up the equation into two cases. The equation is satisfied either if
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<math>\text{cos }x=0\text{ }</math>
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or if
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<math>2\sin x-\sqrt{2}=0</math>.
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<math>\text{cos }x=0\text{ }</math>: this equation has the general solution
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{{Displayed math||<math>x = \frac{\pi}{2}+n\pi\qquad</math>(''n'' is an arbitrary integer).}}
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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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<math>2\sin x-\sqrt{2}=0</math>:
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<math>2\sin x-\sqrt{2}=0</math>: If we collect
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If we collect <math>\sin x</math> on the left-hand side, we obtain the equation
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<math>\text{sin }x</math>
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<math>\sin x = 1/\!\sqrt{2}</math>, which has the general solution
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on the left-hand side, we obtain the equation
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<math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the general solution
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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{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{3\pi }{4}+2n\pi \\
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\end{array} \right.</math>
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(
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<math>n</math>
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an arbitrary integer)
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The complete solution of the equation is
The complete solution of the equation is
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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</math>
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an arbitrary integer).
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Current revision

If we use the formula for double angles, \displaystyle \sin 2x = 2\sin x\cos x, and move all the terms over to the left-hand side, the equation becomes

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

Then, we see that we can take a factor \displaystyle \cos x out of both terms,

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

and hence divide up the equation into two cases. The equation is satisfied either if \displaystyle \cos x = 0 or if \displaystyle 2\sin x-\sqrt{2} = 0\,.


\displaystyle \cos x = 0:

This equation has the general solution

\displaystyle x = \frac{\pi}{2}+n\pi\qquad(n is an arbitrary integer).


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

If we collect \displaystyle \sin x on the left-hand side, we obtain the equation \displaystyle \sin x = 1/\!\sqrt{2}, which has the general solution

\displaystyle \left\{\begin{align}

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

where n is an arbitrary integer.


The complete solution of the equation is

\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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