How to solve #cos(pi/4-x)=sin(pi/6+x)#?

3 Answers
maganbhai P.
Jun 11, 2018

#=>x=kpi+(7pi)/24 ,k inZZ#

Explanation:

We know that,
#color(red)((1)sinC-sinD=2cos((C+D)/2)sin((C-D)/2)#
#color(blue)((2)cos(pi/2-theta)=sintheta#

Here,

#cos(pi/4-x)=sin(pi/6+x)#

#=>cos(pi/4-x)-sin(pi/6+x)=0#

#=>cos[pi/2-pi/4-x]-sin(pi/6+x)#=#0to[becausepi/2-pi/4=pi/4]#

#=>color(blue)(cos[pi/2-(pi/4+x)])-sin(pi/6+x)=0tocolor(blue)(Apply(2)#

#=>color(red)(sin(pi/4+x)-sin(pi/6+x)=0#

Using #color(red)((1) # we get

#color(red)(2cos((pi/4+x+pi/6+x)/2)sin((pi/4+x-pi/6-x)/2)=0#

#=>2cos(((5pi)/12+2x)/2)sin((pi/12)/2)=0#

#=>cos((5pi)/24+x)=0 ...to[because2sin(pi/24)!=0]#

#=>cos(x+(5pi)/24)=0#

#=>color(brown)(x+(5pi)/24=(2k+1)pi/2,k inZZ#

#=>x=kpi+pi/2-(5pi)/24 ,k inZZ#

#=>x=kpi+(7pi)/24 ,k inZZ#

Note:

#color(brown)(costheta=0=>theta=(2k+1)pi/2,k in ZZ#

Nghi N.
Jun 11, 2018

#x = (7pi)/24 + kpi#

Explanation:

#cos (pi/4 - x) = sin (pi/6 + x)#
Reminder:
#sin (pi/6 - x) = cos (pi/2 - (pi/6 + x)) = cos (pi/3 - x)#
Therefor,
#cos (pi/4 - x) = cos (pi/3 - x)#
Trig table, unit circle and property of cos function -->
#pi/4 - x = +- (pi/3 - x)#

a. #pi/4 - x = pi/3 - x# (rejected),
b. #pi/4 - x = - pi/3 + x#
#2x = pi/4 + pi/3 = (7pi)/12 + 2kpi#
#x = (7pi)/24 + kpi#
Check
#x = (7pi)/24 = 52^@5# --> cos (45 - 52.5) = cos (7.5) = 0.99 -->
sin (30 + 52.5) = sin (82.5) = 0.99. Proved.

maganbhai P.
Jun 11, 2018

#=>x=kpi+(7pi)/24 ,k inZZ#

Explanation:

#II^(nd) method :#

We know that,

#color(brown)((1)sintheta=cos(pi/2-theta)#

#color(red)((2)cosC-cosD=-2sin((c+D)/2)sin((C-D)/2)#

#color(blue)((3)sintheta=0=>theta=kpi ,k inZZ#

Here,

#cos(pi/4-x)=color(brown)(sin(pi/6+x)...tocolor(brown)(Apply(1)#

#=>cos(pi/4-x)=color(brown)(cos[pi/2-(pi/6+x)]#

#=>cos(pi/4-x)=cos(pi/2-pi/6-x)#

#=>cos(pi/4-x)=cos(pi/3-x)#

#=>color(red)(cos(pi/4-x)-cos(pi/3-x)=0#

Using #color(red)((1)#,we get

#=>color(red)(-2sin((pi/4-x+pi/3-x)/2)sin((pi/4-x-pi/3+x)/2)=0#

#=>-2sin(((7pi)/12-2x)/2)sin((-pi/12)/2)=0#

#=>sin((7pi)/24-x)=0to[because-2sin(-pi/24)!=0]#

#=>color(blue)(sin(x-(7pi)/24)=0...tocolor(blue)(Apply(3)#

#=>color(blue)(x-(7pi)/24=kpi, k inZZ#

#=>x=kpi+(7pi)/24 ,k inZZ#

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