How do you solve $sqrt2cosxsinx-cosx=0$ ?

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Answer 1 · 2016-11-16T16:12:08

The solution is $S={pi/2,(5pi)/2,pi/4,(3pi)/4}$

Let's factorise the equation

$sqrt2cosxsinx-cosx=0$

$cosx(sqrt2sinx -1)=0$

Therefore, $cosx=0$ or $sinx=1/sqrt2$

We are looking for $x in [0,2pi]$

For, $cosx=0$ $=>$ $x=pi/2$ and $x=(5pi)/2$

For $sinx=1/sqrt2$ $=>$ , $x=pi/4$ and $x=(3pi)/4$

Answer 2 · 2016-11-16T16:23:44

Factor the expression as follows

$cosx*(sqrt2*sinx-1)=0$

Hence now you have that

$cosx=0$ or $sqrt2*sinx-1=0$

The general solution of $cosx=0$ is $x=2*k*pi+-pi/2$

where $k$ is an integer.

The gneral solution of $sqrt2*sinx-1=0$ or $sinx=1/sqrt2=sqrt2/2$

is $x=2*n*pi+pi/4$ or $x=2*n*pi+(3*pi)/4$

where $n$ is an integer.

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