How do you solve $2sinx+cscx=3$ ?

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Answer 1 · 2016-12-11T10:40:35

The solutions are $S={pi/2+2kpi,pi/6+2kpi,(5pi)/6+2kpi}$ , $kinZZ$

$cscx=1/sinx$

We rewrite the equation

$2sinx+1/sinx=3$

$2sin^2x+1=3sinx$

$2sin^2x-3sin+1=0$

This is a quadratic equation in $sinx$

$ax^2+bx+c=0$

We calculate the discriminant

$Delta=b^2-4ac=9-4*2*1=1$

As, $Delta>0$ , we have 2 real roots

The roots are

$(-b+-sqrtDelta)/(2a)$

so,
$sinx=(3+-1)/4$

$sinx=1$ , $=>$ , $x=pi/2+2kpi$ , $(k in ZZ)$

$sinx=1/2$ , $=>$ , $x=pi/6+2kpi and$ $x=(5pi)/6+2kpi$
$k in ZZ$

How do you solve 2sinx+cscx=32sinx+cscx=3?