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,

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

$=>8costheta-5costheta=1$

$=>3costheta=1$

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

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

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

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

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

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

How do you find the solution to 5costheta+1=8costheta5costheta+1=8costheta if 0<=theta<2pi0<=theta<2pi?