#1/costheta+3sinthetatantheta+4=0# can be expressed as #3cos^2theta - 4costheta -4=0#. Hence solve the equation #1/costheta+3sinthetatantheta+4=0# for #0<=theta<=360# ?

2 Answers
Jim G.
Nov 23, 2017

#x=131.8^@" or "x=228.2^@#

Explanation:

#1/costheta+3sinthetaxxsintheta/costheta+4=0#

#rArr1/costheta+(3sin^2theta)/costheta+(4costheta)/costheta=0#

#rArr(1+3(1-cos^2theta)+4costheta)/costheta=0#

#rArr(-3cos^2theta+4costheta+4)/costheta=0"#

#rArr-3cos^2theta+4costheta+4=0larrcolor(blue)"multiply by - 1"#

#rArr3cos^2theta-4costheta-4=0#

#"using "3cos^2theta-4costheta-4=0" to solve"#

#3cos^2theta+2costheta-6costheta-4=0larr"split "costheta" term"#

#color(red)(costheta)(3costheta+2)color(red)(-2)(3costheta+2)=0larr"grouping"#

#rArr(costheta+2)(color(red)(3costheta+2))=0#

#"equate each factor to zero and solve for "theta#

#costheta+2=0tocostheta=-2larrcolor(red)"no solution"#

#3costheta+2=0tocostheta=-2/3#

#costheta<0rArrthetacolor(blue)" in second/third quadrants"#

#theta=cos^-1(2/3)=48.2^@larrcolor(red)"related acute angle"#

#rArrtheta=(180-48.2)^@=131.8^@#

#rArrtheta=(180+48.2)^@=228.2^@#

P dilip_k
Nov 23, 2017

#1/costheta+3sinthetatantheta+4=0#

#=>1/costheta+3sintheta*sintheta/costheta+4=0#

#=>1+3sin^2theta+4costheta=0#

#=>1+3(1-cos^2theta)+4costheta=0#

#=>3cos^2theta-4costheta-4=0#

#=>3cos^2theta-6costheta+2costheta-4=0#

#=>3costheta(costheta-2)+2(costheta-2)=0#

#=>(3costheta+2)(costheta-2)=0#

#costheta=2# not possible

So #costheta=-2/3=cos131.8^@=cos(360-131.8)^@#

Hence

#theta=131.8^@ and theta=228.2^@#

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