Lösung 4.4:8a
Aus Online Mathematik Brückenkurs 1
K (Lösning 4.4:8a moved to Solution 4.4:8a: Robot: moved page) |
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- | {{ | + | If we use the formula for double angles, |
- | < | + | <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 |
- | {{ | + | |
- | {{ | + | |
- | < | + | <math>2\sin x\cos x-\sqrt{2}\cos x=0.</math> |
- | {{ | + | |
+ | |||
+ | Then, we see that we can take a factor cos x out of both terms, | ||
+ | |||
+ | |||
+ | <math>\cos x\left( 2\sin x-\sqrt{2} \right)=0</math> | ||
+ | |||
+ | |||
+ | and hence divide up the equation into two cases. The equation is satisfied either if | ||
+ | <math>\text{cos }x=0\text{ }</math> | ||
+ | or if | ||
+ | <math>2\sin x-\sqrt{2}=0</math>. | ||
+ | |||
+ | |||
+ | <math>\text{cos }x=0\text{ }</math>: this equation has the general solution | ||
+ | |||
+ | |||
+ | <math>x=\frac{\pi }{2}+n\pi </math> | ||
+ | ( | ||
+ | <math>n</math> | ||
+ | an arbitrary integer) | ||
+ | |||
+ | |||
+ | <math>2\sin x-\sqrt{2}=0</math>: If we collect | ||
+ | <math>\text{sin }x</math> | ||
+ | on the left-hand side, we obtain the equation | ||
+ | <math>\text{sin }x\text{ }={1}/{\sqrt{2}}\;</math>, which has the general solution | ||
+ | |||
+ | |||
+ | <math>\left\{ \begin{array}{*{35}l} | ||
+ | x=\frac{\pi }{4}+2n\pi \\ | ||
+ | x=\frac{3\pi }{4}+2n\pi \\ | ||
+ | \end{array} \right.</math> | ||
+ | ( | ||
+ | <math>n</math> | ||
+ | an arbitrary integer) | ||
+ | |||
+ | The complete solution of the equation is | ||
+ | |||
+ | |||
+ | <math>\left\{ \begin{array}{*{35}l} | ||
+ | x=\frac{\pi }{4}+2n\pi \\ | ||
+ | x=\frac{\pi }{2}+n\pi \\ | ||
+ | x=\frac{3\pi }{4}+2n\pi \\ | ||
+ | \end{array} \right.</math> | ||
+ | ( | ||
+ | <math>n</math> | ||
+ | an arbitrary integer). |
Version vom 13:31, 1. Okt. 2008
If we use the formula for double angles, \displaystyle \text{sin 2}x=\text{2sin }x\text{ 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.
Then, we see that we can take a factor cos x out of both terms,
\displaystyle \cos x\left( 2\sin x-\sqrt{2} \right)=0
and hence divide up the equation into two cases. The equation is satisfied either if
\displaystyle \text{cos }x=0\text{ }
or if
\displaystyle 2\sin x-\sqrt{2}=0.
\displaystyle \text{cos }x=0\text{ }: this equation has the general solution
\displaystyle x=\frac{\pi }{2}+n\pi
(
\displaystyle n
an arbitrary integer)
\displaystyle 2\sin x-\sqrt{2}=0: If we collect
\displaystyle \text{sin }x
on the left-hand side, we obtain the equation
\displaystyle \text{sin }x\text{ }={1}/{\sqrt{2}}\;, which has the general solution
\displaystyle \left\{ \begin{array}{*{35}l}
x=\frac{\pi }{4}+2n\pi \\
x=\frac{3\pi }{4}+2n\pi \\
\end{array} \right.
(
\displaystyle n
an arbitrary integer)
The complete solution of the equation is
\displaystyle \left\{ \begin{array}{*{35}l}
x=\frac{\pi }{4}+2n\pi \\
x=\frac{\pi }{2}+n\pi \\
x=\frac{3\pi }{4}+2n\pi \\
\end{array} \right.
(
\displaystyle n
an arbitrary integer).