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#