How do you solve $8sinxcosx + 4 = 0$ ?

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Answer 1 · 2016-05-29T17:31:20

Using the formula for $sin(2x)$ .

There is a trigonometric formula that tells you that

$sin(2x)=2sin(x)cos(x)$

so we can use it in the problem dividing everything by 4.

$(8sin(x)cos(x))/4+4/4=0$
$2sin(x)cos(x)+1=0$
$sin(2x)=-1$ .

We know that the $sin$ is equal to $-1$ when the angle is $3/2pi+2npi$ where n can be any integer number. So we have

$2x=3/2pi+2npi$
$x=3/4pi+npi$

How do you solve 8sinxcosx + 4 = 08sinxcosx + 4 = 0?