If sin square theta-cos square theta=2-5cos theta then what is the value of theta?

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Answer 1 · 2018-08-12T11:55:04

#theta=2kpi+-arc cos((5-sqrt17)/4) ,k inZZ#

Here ,

#sin^2theta-cos^2theta=2-5costheta ,.......0<= theta <=2pi#

#:.1-cos^2theta-cos^2theta=2-5costheta#

#:.2cos^2theta-5costheta+1=0#

Comparing with #ax^2+bx+c=0#

#a=2 , b=-5 andc=1#

#Delta=b^2-4ac=(-5)^2-4(2)(1)=17#

So ,

#costheta=(-b+-sqrtDelta)/(2a)=(5+-sqrt17)/(2*2)#

#:.costheta=(5-sqrt17)/4 or costheta=(5+sqrt17)/4 #

#:.costheta~~0.22 or costheta~~2.28 !in [-1,1]#

#:.costheta~~0.22 >0=>1^(st)Quadrant or 4^(th)Quadrant#

#:.theta=arc cos((5-sqrt17)/4) or theta=2pi-arc cos((5- sqrt17)/4)#

#:.theta=(1.35)^R # OR # theta=2pi-(1.35)^R# ,where , #0 <= theta < 2pi#

The general solution Is :

#theta=2kpi+-arc cos((5-sqrt17)/4) ,k inZZ#