Solve the equation cos5x = sinx ?

We know that,

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

#color(violet)((2)costheta=cosalpha => theta= 2kpi +-alpha, kinZZ#

Here,

#cos5x = sinx# , where #0 <= x <= 360^circ#

#=>cos5x=cos(pi/2-x) ,where, x in[0, 2pi]#

#=>5x=2kpi+-(pi/2-x),kinZZ...to color(violet)(Apply (2)#

#=>5x=2kpi+pi/2-x or 5x=2kpi-pi/2+x, kinZZ#

We have two options :

#(i)5x=2kpi+pi/2-x, kinZZ#

#=>5x+x=2kpi+pi/2, kinZZ#

#=>6x=(4k+1)pi/2 ,kinZZ#

#=>x=(4k+1)pi/12 ,kinZZ#

#k=0=>color(red)(x=pi/12 and pi/12in [0,2pi]#

#k=1=>color(red)(x=(5pi)/12 and (5pi)/12 in [0, 2pi]#

#k=2=>color(red)(x=(9pi)/12 and (9pi)/12 in [0, 2pi]#

#k=3=>color(red)(x=(13pi)/12 and (13pi)/12 in [0, 2pi]#

#k=4=>color(red)(x=(17pi)/12 and (17pi)/12 in [0, 2pi]#

#k=5=>color(red)(x=(21pi)/12 and (21pi)/12 in [0, 2pi]#

#k=6=>color(blue)(x=(25pi)/12 and (25pi)/12 !in [0, 2pi]#

#(ii)5x=2kpi-pi/2+x, kinZZ#

#=>5x-x=2kpi-pi/2, kinZZ#

#=>4x=(4k-1)pi/2 ,kinZZ#

#=>x=(4k-1)pi/8 ,kinZZ#

#k=0=>color(blue)(x=-pi/8 and -pi/8 !in[0, 2pi]#

#k=1=>color(red)(x=(3pi)/8 and (3pi)/8 in[0,2pi]#

#k=2=>color(red)(x=(7pi)/8 and (7pi)/8 in[0,2pi]#

#k=3=>color(red)(x=(11pi)/8 and (11pi)/8 in[0,2pi]#

#k=4=>color(red)(x=(15pi)/8 and (15pi)/8 in[0,2pi]#

#k=5=>color(blue)(x=(19pi)/8 and (19pi)/8 !in[0,2pi]#

Hence,

#x=pi/12,(5pi)/12,(9pi)/12,(13pi)/12, (17pi)/12, (21pi)/12,(3pi)/8,(7pi)/8,(11pi)/8,(15pi)/8.#

OR

#x=15^circ,75^circ,135^circ,195^circ,255^circ,315^circ,67.5^circ,157.5^circ,247.5^circ,337.5^circ#

Write your equation in the form
#cos(5x)-sin(x)=-2(2sin(2x)-1)*sin(pi/4+x)*sin(pi/4+2x)=0#
and set the single factors equal to Zero.
And for the seond Problem write
#sin(2x)-cos(2x)=-2sin(pi/4-x/2)^2(-1+2cos(2x)-2sin(x))#