How to solve 3sin(x)=cos^2(x)?

#3sin(x)=2cos^2(x)#?

2 Answers
Douglas K.
May 13, 2018

Given: #3sin(x)=2cos^2(x)#

Divide both sides of the equation by 2:

#3/2sin(x)=cos^2(x)#

Substitute: #cos^2(x) = 1-sin^2(x)#:

#3/2sin(x)=1-sin^2(x)#

Add #sin^2(x)-1# to both sides:

#sin^2(x)+3/2sin(x)-1=0#

Please observe that the above is a quadratic equation where sin(x) is the variable, therefore, we can use the quadratic formula:

#sin(x) = (-3/2+-sqrt((3/2)^2-4(1)(-1)))/2#

#sin(x) = 1/2#

#sin(x) = -2 larr# discard because it is outside of the domain of the sine function.

#sin(x) = 1/2#

#x = sin^-1(1/2)#

The values for this condition are well known:

#x = pi/6# and #x = (5pi)/6#

This repeats at integer multiples of #2pi#:

#x = pi/6+2npi# and #x = (5pi)/6+ 2npi# #n in ZZ#

maganbhai P.
May 13, 2018

#x=kpi+(-1)^k*pi/6,kinZZ#

Explanation:

Here,

#3sinx=2cos^2x#

#3sinx=2(1-sin^2x)#

#3sinx=2-2sin^2x#

#2sin^2x+3sinx-2=0#

#2sin^2x+4sinx-sinx-2=0#

#2sinx(sinx+2)-1(sinx+2)=0#

#(2sinx-1)(sinx+2)=0#

#2sinx-1=0 or sinx+2=0#

#2sinx=1 or sinx=-2 !in [-1,1]#

So,

#sinx=1/2#

#sinx=sin(pi/6)#

#x=kpi+(-1)^k*pi/6,kinZZ#

Note:

If #x in[0,2pi),then#

#sinx=1/2 > 0=>I^(st)Quadrant or II^(nd)Quadrant#

#=>x=pi/6 or x=pi-pi/6=(5pi)/6#

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