How do you find the solution to #5costheta+1=8costheta# if #0<=theta<2pi#?

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Answer 1 · 2018-05-09T18:44:09

#theta=cos^-1(1/3) or theta=2pi-cos^-1(1/3)#

Here,

#5costheta+1=8costheta,where, 0 <= theta <= 2pi#

#=>8costheta-5costheta=1#

#=>3costheta=1#

#=>costheta=1/3 > 0=>I^(st) Quadrant orIV^(th) Quadrant#

#=>theta =cos^-1(1/3)#

#(i)I^(st)Quadrant=> 0 <= theta <= pi/2#

#=>theta=cos^-1(1/3)=>theta~~1.23^R or theta~~(78.37)^circ#

#(ii)IV^(th)Quadrant=>(3pi)/2 <= theta <=2pi#

#=>theta=2pi-cos^-1(1/3)~~5.05^R or theta~~(281.63)^circ#