How do you solve $Cos(2x)cos(x)-sin(2x)sin(x)=0$ ?

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Answer 1 · 2016-07-06T15:01:50

$x = pi/6 pm k pi/3, {k = 0,pm1,pm2,pm3,...}$

Using $Cos(a + b)=Cos(a) Cos(b) - Sin(a)Sin(b)$
$Cos(2x)cos(x)-sin(2x)sin(x) = cos(3x) = 0$

So

$3x=pi/2pm2k pi, {k = 0,pm1,pm2,pm3,...}$

Finally

$x = pi/6 pm k pi/3, {k = 0,pm1,pm2,pm3,...}$

How do you solve Cos(2x)cos(x)-sin(2x)sin(x)=0Cos(2x)cos(x)-sin(2x)sin(x)=0?