How do you solve #3sintheta-4costheta=1#?

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Answer 1 · 2016-10-11T18:14:20

#theta = 221.5931434^@+360n^@# OR #theta=64.66706143^@+360n^@#

#3sin theta-4cos theta=1#

Use : #Acosx+Bsinx=Ccos(x-D)# where #C=sqrt(A^2+B^2) and D=cos^-1 (A/C)#

#A=-4,B=3, -> QII#
# C=sqrt(16+9)=sqrt25 = 5#

#D=cos^-1(-4/5)~~143.13^@#

#5cos(theta-143.1301024^@)=1#

#cos(theta-143.1301024^@)=1/5#

#(theta-143.1301024^@)=cos^-1(1/5)#

#theta-143.1301024^@=+-78.46304097^@+360n^@#

#theta=+-78.46304097^@+360n^@ + 143.1301024^@#

#theta = 221.5931434^@+360n^@# OR #theta=64.66706143^@+360n^@#