How do you solve $cosx(3sinx - 5) = 0$ ?

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Answer 1 · 2016-02-01T12:35:23

$(2n+1)pi/2$ [where $n in ZZ$ ]

here,
$cosx(3sinx-5)=0$

$or,3sinxcosx-5cosx=0$

$or,3*2sinxcosx=5*2cosx$

$or,3sin2x=10cosx$

$or,9sin^2 2x=100cos^2x$ [taking square on each sides]

$or,9*2sin^2 2x=100*2cos^2x$

$or,18(1-cos^2 2x)=100(1+cos2x)$

$or,18-18cos^2 2x=100+100cos2x$

$or,18cos^2 2x+100cos2x+82=0$

$or, 18cos^2 2x+18cos2x+82cos2x+82=0$

$or,18cos2x(cos2x+1)+82(cos2x+1)=0$

$or,(cos2x+1)(18cos2x+82)=0$

one solution is,

$cos2x+1=0$

$or,cos2x=-1$

$or,2x=(2n+1)pi$ [where, $ninZZ$ ]

$or,x=(2n+1)pi/2$

another solution is,

$18cos2x+82=0$

$or,cos2x=-82/18$

but, $cos2x!=-82/18$ [as, $cos2x cancel(>)1$ ]

so, $x=(2n+1)pi/2$ [where $n in ZZ$ ]

How do you solve cosx(3sinx - 5) = 0cosx(3sinx - 5) = 0?