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

1 Answer
Feb 1, 2016

Answer:

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

Explanation:

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#]