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

3 Answers
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

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.

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