How do you solve $sin2xsinx-cosx=0$ ?

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How do you solve $sin2xsinx-cosx=0$ ?

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Answer 1 · 2016-11-08T16:49:17

The values of $x in { pi/2,(3pi)/2,pi/4,(3pi)/4,(5pi)/4,(7pi)/4}$

Explanation:

$Sin2x=2sinxcosx$
$sin2xcosx-cosx=2sinxcosxsinx-cosx=cosx(2sin^2x-1)$
$cosx=0$ and $2sin^2x-1=0$ $=>$ $sinx=+-1/sqrt2$
Let the solutions $x in[0,2pi]$
The solution of the first equation is $(pi/2,(3pi)/2)$
The solution of the second equations are $(pi/4,(3pi)/4,(5pi)/4,(7pi)/4)$

Answer 2 · 2016-11-08T17:17:08

$x=(kpi)/2 or x=(kpi)/4 k in ZZ$

Explanation:

Solving the given trigonometric equation $sin2xsinx - cosx$ is determined by:

Substituting some trigonometric identities.

$color(blue)(sin2x=2sinxcosx)$

$color(red)(cos2x=1 - 2sin^2)$

Performing some calculations.

Factorizing $sin2x - cosx$

$" "$

Solving the given equation:

$sin2xsinx-cosx=0$

$rArr(color(blue)(2sinxcosx))sinx-cosx=0$

$rArr2sin^2xcosx-cosx=0$

$rArrcosx(color(red)(2sin^2x-1))=0$

$rArrcosx(color(red)(-cos2x))=0$

$rArr-cosxcos2x=0$

$cosx =0 rArr x=(kpi)/2" " k in ZZ$

Or

$cos2x=0$
$rArr 2x=(kpi)/2$
$rArr x=(kpi)/4 " " k in ZZ$

Therefore,

$x=(kpi)/2 or x=(kpi)/4 k in ZZ$

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