Question #2898e

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Answer 1 · 2016-08-30T01:49:38

$sin2x=sqrt3sinx$

$=>2sinxcosx-sqrt3sinx=0$

$=>sinx(2cosx-sqrt3)=0$

This gives $" "sinx=0 and cosx=sqrt3/2$

When $sinx=0$
the general suolution is

$x=npi," where "n in ZZ$

Since $" "0<=x<=360$

The solution within this limit can be had for $n =0 and 1$

So $" "x = 0 or pi$

Again when $cosx=sqrt3/2=cos(pi/6)$
the general suolution is

$x=2npi+-pi/6," where "n in ZZ$

Since $" "0<=x<=360$

The solution within this limit can be had for $n =0 and 1$

For n=0 $" "x = pi/6=30^@$

For n=1 $" "x = 2pi-pi/6=(11pi)/6=330^@$