How do you solve $2cos2x+3=2$ for $0<=x<=360$ ?

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Answer 1 · 2016-08-28T02:31:57

$x = 60^o, 120^o, 240^o and 300^o$ ..

Here,

$cos 2x$

$= (2-3)/2$

$=-1/2$

$=-cos 60^o$

$=cos(180^o-60^o)$

$=cos(120^o$ ,

and so,

the principal vale of $2x = 120^o$

The general value is

$(2nX180+-120)^o, n = 0, +-1, +-2, +-3,$ , giving,

$x=(nX180+-60)^o, n=0, +-1, +-2, +-3, ...$

For the range $(0^o, 360^o)$ , choose + sign and, n = 0, and 1, and

for - sign, n=1 and 2.

Combining and ordering, ,

$x = 60^o, 120^o, 240^o and 300^o$ .

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