How do you solve #\sin ( x + 4) = \cos ( 2x - 1)#?

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Answer 1 · 2018-02-24T20:26:47

#x={(-1+1/6pi)+2/3kpi,(5-1/2pi)-2kpi|k in ZZ}#

#sin(x+4)=cos(2x-1)#

Use #cosA=sin(pi/2-A)# to write the equation in sines.

#sin(x+4)=sin (pi/2-(2x-1))=sin(1+pi/2-2x)#

Take the #arcsin# on both sides and remember that #sinx=sin(pi-x)#

#x+4=1+pi/2-2x+2kpi#, #kinZZ#

#x+4=-1+1/2pi+2x+2kpi, k in ZZ#

Solve each equation separately

#x+4=1+pi/2-2x+2kpi#, #kinZZ#

#3x=(-3+1/2pi)+2kpi#

#x=(-1+1/6pi)+2/3kpi#

#x+4=-1+1/2pi+2x+2kpi, k in ZZ#

#x=(5-1/2pi)-2kpi#