Question #071e3

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Question #071e3

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Answer 1 · 2017-04-08T00:39:10

$x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ$

Explanation:

If $ab=0$ , then either $a=0$ or $b=0$ or $a=b=0$ .

Looking at the equation $(sin(x)+3)(2sin(x)+1)=0$ , it can be said that the solutions occur when $sin(x)+3=0$ or $2sin(x)+1=0$ .

Consider $sin(x)+3=0$ first. The range of $sin(x)$ is $-1≤sin(x)≤1$ for all $x in RR$ . There are no solutions here.

Consider $2sin(x)+1=0$ . This is equivalent to $sin(x)=-1/2$ . Solving for $x$ in the range $[0,2pi)$ , we get $(7pi)/6$ and $(11pi)/6$ .

Since $sin(x)$ is periodic with period $2pi$ , the solution is $x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ$ .

Answer 2 · 2017-04-08T00:42:31

$x=(7pi)/6+2kpi" or "x=(11pi)/6+2kpi$
$color(white)("XXXXXXXXXXXXXXXXXXXXX")AAk in ZZ$

Explanation:

If $(sin(x)+3)(2sin(x)+1)=0$
${: ("either ",sin(x)+3=0," or ",2sin(x)+1=0), (,"impossible",,rarr sin(x)=-1/2), (,color(white)("X")"since "sin(x) in[-1,+1],,x=(7pi)/6" or "(11pi)/6), (,,,color(white)("X")"assuming "x in [0,2pi)) :}$

Detail:
$color(white)("XXX")sin(x)=-1/2$
implies a reference angle of $pi/6$ (one of the standard angles)
in either Quadrant III or IV

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Question #071e3

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