How do you solve #3cscA-2sinA-5=0#?

2 Answers
maganbhai P.
Mar 4, 2018

#A=kpi+(-1)^k(pi/6),kinZ#

Explanation:

#3cscA-2sinA-5=0#
#rArr3/sinA-2sinA-5=0#
#rArr3-2sin^2A-5sinA=0#
#rArr2sin^2A+5sinAcolor(red)(-3)=0#
#rArr2sin^2A+6sinA-sinA-3=0#
#rArr2sinA(sinA+3)-1(sinA+3)=0#
#rArr(sinA+3)(2sinA-1)=0#
#rArrsinA=-3!in[-1,1],sinA=1/2in[-1,1]#
#rArrsinA=sin(pi/6)#
#rArrA=kpi+(-1)^k(pi/6),kinZ#
#rArrA=kpi+(-1)^k(pi/6),kinZ#

1s2s2p
Mar 4, 2018

#A=(npi)/2+-pi/3, ninZZ#
#color(white)(A)=n90^circ+-60^circ, ninZZ#

Explanation:

First we multiply everything by #sinA# since #cscA=1/sinA# and sinA/sinA=1#

#sinA(3cscA-2sinA-5)=sinA(0)#

#3-2sin^2A-5sinA=0#

Substitute #x=sinA#

#2x^2+5x-3=0#

#x=(-b+-sqrt(b^2-4ac))/2#

#color(white)(x)=(-5+-sqrt(5^2-4(2*-3)))/4#

#color(white)(x)=(-5+-sqrt(25+24))/4#

#color(white)(x)=(-5+-sqrt(49))/4#

#color(white)(x)=(-5+-7)/4#

#color(white)(x)=(-5-7)/4 or (-5+7)/4#

#color(white)(x)=-12/4 or 2/4#

#color(white)(x)=-3 or 1/2#

However, #-1lesinAle1# so we must take #1/2#

#sinA=1/2#

#A=arcsin(1/2)=pi/6-=30^circ, A=(5pi)/6-=150^circ#

#A=(npi)/2+-pi/3, ninZZ#
#color(white)(A)=n90^circ+-60^circ, ninZZ#

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