How do you solve #sqrt2cosxsinx-cosx=0#?

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Answer 1 · 2016-11-16T16:12:08

The solution is #S={pi/2,(5pi)/2,pi/4,(3pi)/4} #

Let's factorise the equation

#sqrt2cosxsinx-cosx=0#

#cosx(sqrt2sinx -1)=0#

Therefore, #cosx=0# or #sinx=1/sqrt2#

We are looking for #x in [0,2pi]#

For, #cosx=0# #=>##x=pi/2# and #x=(5pi)/2#

For #sinx=1/sqrt2# #=>#, #x=pi/4# and #x=(3pi)/4#

Answer 2 · 2016-11-16T16:23:44

Factor the expression as follows

#cosx*(sqrt2*sinx-1)=0#

Hence now you have that

#cosx=0# or #sqrt2*sinx-1=0#

The general solution of #cosx=0# is #x=2*k*pi+-pi/2#

where #k# is an integer.

The gneral solution of #sqrt2*sinx-1=0# or #sinx=1/sqrt2=sqrt2/2#

is #x=2*n*pi+pi/4# or #x=2*n*pi+(3*pi)/4#

where #n# is an integer.