Question #fce84

1 Answer
Jun 8, 2016

Answer:

#x=pi/2+kpi,pi/4+(kpi)/2,k in ZZ#

Explanation:

Use the identity #sin2x=2sinxcosx# to rewrite the expression.

#sin2xsinx=cosx" "=>" "2sinxcosx(sinx)=cosx#

#=>" "2sin^2xcosx=cosx#

Subtract #cosx# from each side.

#2sin^2xcosx-cosx=0#

Factor #cosx#.

#cosx(2sin^2x-1)=0#

Set both of these terms equal to #0#.

#cosx=0#

This occurs at #x=pi/2# and #x=(3pi)/2# on the interval #[0,2pi)#, but can be generalized as #x=pi/2+kpi# where #k# is an integer.

The other term is

#2sin^2x-1=0#

#sin^2x=1/2#

#sinx=+-sqrt(1/2)=+-1/sqrt2=+-sqrt2/2#

Sine equals #+-sqrt2/2# at #x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4# on #[0,2pi)#, which can be generalized to #pi/4+(kpi)/2#, where #k# is an integer.

Note that the mathematical way to express that #k# is an integer is to say that #k in ZZ#.