Answer 4.4:2
From Förberedande kurs i matematik 1
(Difference between revisions)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <math>\left\{\eqalign{ x&=\displaystyle\frac{\pi}{3}+2n\pi\cr x&=\displaystyle\frac{2\pi}{3}+2n\pi } \right.</math> |b) |width="33%" ...) |
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|width="33%" | <math>\left\{\eqalign{ | |width="33%" | <math>\left\{\eqalign{ | ||
x&=\displaystyle\frac{\pi}{4}+\displaystyle\frac{2n\pi}{3}\cr | x&=\displaystyle\frac{\pi}{4}+\displaystyle\frac{2n\pi}{3}\cr | ||
| - | x&=\displaystyle\frac{5\pi}{12}+\displaystyle\frac{2n\pi}{3}} \right. | + | x&=\displaystyle\frac{5\pi}{12}+\displaystyle\frac{2n\pi}{3}} \right.</math> |
|} | |} | ||
Revision as of 12:52, 3 April 2008
| a) | \displaystyle \left\{\eqalign{
x&=\displaystyle\frac{\pi}{3}+2n\pi\cr x&=\displaystyle\frac{2\pi}{3}+2n\pi } \right. | b) | \displaystyle \left\{\eqalign{
x&=\displaystyle\frac{\pi}{3}+2n\pi\cr x&=\displaystyle\frac{5\pi}{3}+2n\pi } \right. | c) | \displaystyle x=n\pi |
| d) | \displaystyle \left\{\eqalign{
x&=\displaystyle\frac{\pi}{20}+\displaystyle\frac{2n\pi}{5}\cr x&=\displaystyle\frac{3\pi}{20}+\displaystyle\frac{2n\pi}{5} } \right. | e) | \displaystyle \left\{\eqalign{
x&=\displaystyle\frac{\pi}{30}+\displaystyle\frac{2n\pi}{5}\cr x&=\displaystyle\frac{\pi}{6}+\displaystyle\frac{2n\pi}{5}} \right. | f) | \displaystyle \left\{\eqalign{
x&=\displaystyle\frac{\pi}{4}+\displaystyle\frac{2n\pi}{3}\cr x&=\displaystyle\frac{5\pi}{12}+\displaystyle\frac{2n\pi}{3}} \right. |
