What are the exact solutions for 2cosϴ+2sinϴ=√ in [0,2π) interval?

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Answer 1 · 2018-04-07T20:39:41

#theta=75.93^@, 104.7^@#

#theta=15.07^@, 164.93^@#

.

#2costheta+2sintheta=sqrt6#

#costheta+sintheta=sqrt6/2#, #color(red)(Equation-1)#

#(costheta+sintheta)^2=6/4=3/2#

#cos^2theta+sin^2theta+2sinthetacostheta=3/2#

#1+2sinthetacostheta=3/2#

#2sinthetacostheta=3/2-1=1/2#

#sinthetacostheta=1/4#

From #color(red)(Equation-1)#:

#costheta=sqrt6/2-sintheta#

Let's plug this in:

#sintheta(sqrt6/2-sintheta)=1/4#

#sqrt6/2sintheta-sin^2theta=1/4#

#sin^2theta-sqrt6/2sintheta+1/4=0#

Let's multiply the equation by #4#:

#4sin^2theta-2sqrt6sintheta+1=0#

Let #sintheta=x#

#4x^2-2sqrt6x+1=0#

Using the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#:

#x=(2sqrt6+-sqrt(24-4(4)(1)))/(2(4))=(2sqrt6+-sqrt8)/8=(2sqrt6+-2sqrt2)/8#

#x=(sqrt6+-sqrt2)/4#

#x=0.97, 0.26#

#sintheta=0.97, :. theta=arcsin(0.97), :. theta=75.93^@ and 104.7^@#

#sintheta=0.26, :. theta=arcsin(0.26), :. theta=15.07^@ and 164.93^@#